Engineering Statics: Open and Interactive

This solution demonstrates solving integrals using vertical rectangular strips. Set the slider on the diagram to \(h\;dx\) to see a representative element.

1. Set up the integrals.

   The bounding functions in this example are vertical lines \(x=0\) and \(x=a\text{,}\) and horizontal lines \(y = 0\) and \(y = h\text{.}\)

   The strip extends from \((x,0)\) on the \(x\) axis to \((x,h)\) on the top of the rectangle, and has a differential width \(dx\text{.}\)

   The area of the strip is the base times the height, so

   \begin{equation*} dA = h\ dx\text{.} \end{equation*}

   The centroid of the strip is located at its midpoint so, by inspection

   \begin{align*} \bar{x}_{\text{el}} \amp = x \\ \bar{y}_{\text{el}} \amp = h/2 \end{align*}

   With vertical strips the variable of integration is \(x\text{,}\) and the limits on \(x\) run from \(x=0\) at the left to \(x=b\) on the right. For a rectangle, both 0 and \(h\) are constants, but in other situations, \(\bar{y}_{\text{el}}\) and the left or right limits may be functions of \(x\text{.}\)

2. Solve the integrals.

   Substitute \(dA\text{,}\) \(\bar{x}_{\text{el}}\text{,}\) and \(\bar{y}_{\text{el}}\) into (7.7.2) and integrate.

   \begin{align*} A \amp = \int dA \amp Q_x \amp = \int \bar{y}_{\text{el}}\ dA \amp Q_y \amp = \int \bar{x}_{\text{el}}\ dA \\ \amp = \int_0^b h\ dx \amp \amp = \int_0^b \frac{h}{2} ( h\ dx ) \amp \amp = \int_0^b x\; (h\ dx)\\ \amp = \Big [ hx \Big ]_0^b \amp \amp = \frac{h^2}{2} \int_0^b dx \amp \amp = h \int_0^b x \ dx\\ \amp = hb - 0 \amp \amp = \frac{h^2}{2} \Big [x \Big ]_0^b \amp \amp = h \left[\frac{x^2}{2} \right ]_0^b\\ A \amp = bh \amp Q_x \amp = \frac{h^2 b}{2} \amp Q_y \amp = \frac{b^2 h}{2} \end{align*}

   Unsurprisingly, we learn that the area of a rectangle is base times height. Since the area formula is well known, it was not really necessary to solve the first integral. Note that \(A\) has units of \([\text{length}]^2\text{,}\) and \(Q_x\) and \(Q_y\) have units of \([\text{length}]^3\text{.}\)

3. Find the centroid.

   Substituting the results into the definitions gives

   \begin{align*} \bar{x} \amp = \frac{Q_y}{A} \amp \bar{y} \amp = \frac{Q_x}{A}\\ \amp = \frac{b^2h}{2} \bigg/ { bh} \amp \amp = \frac{h^2b}{2} \bigg/ { bh}\\ \amp = \frac{b}{2}\amp \amp = \frac{h}{2}\text{.} \end{align*}

This solution demonstrates solving integrals using horizontal rectangular strips. Set the slider on the diagram to \(b\;dy\) to see a representative element.

1. Set up the integrals.

   The bounding functions \(x=0\text{,}\) \(x=a\text{,}\) \(y = 0\) and \(y = h\text{.}\)

   The strip extends from \((0,y)\) on the \(y\) axis to \((b,y)\) on the right, and has a differential height \(dy\text{.}\)

   The area of the strip is the base times the height, so

   \begin{equation*} dA = b\ dy\text{.} \end{equation*}

   The centroid of the strip is located at its midpoint so, by inspection

   \begin{align*} \bar{x}_{\text{el}} \amp = b/2 \\ \bar{y}_{\text{el}} \amp = y \end{align*}

   With horizontal strips the variable of integration is \(y\text{,}\) and the limits on \(y\) run from \(y=0\) at the bottom to \(y = h\) at the top.

   For a rectangle, both 0 and \(h\) are constants, but in other situations, \(\bar{x}_{\text{el}}\) and the upper or lower limits may be functions of \(y\text{.}\)

2. Solve the integrals.

   Substitute \(dA\text{,}\) \(\bar{x}_{\text{el}}\text{,}\) and \(\bar{y}_{\text{el}}\) into (7.7.2) and integrate. The results are the same as we found using vertical strips.

   \begin{align*} A \amp = \int dA \amp Q_x \amp = \int \bar{y}_{\text{el}}\ dA \amp Q_y \amp = \int \bar{x}_{\text{el}}\ dA \\ \amp = \int_0^h b\ dy \amp \amp = \int_0^h y\ ( b\ dy ) \amp \amp = \int_0^h \frac{b}{2} (b\ dy)\\ \amp = \Big [ by \Big ]_0^h \amp \amp = b\int_0^h y\ dy \amp \amp = \frac{b^2}{2} \int_0^h dy\\ \amp = bh \amp \amp = b\ \Big [\frac{y^2}{2} \Big ]_0^h \amp \amp = \frac{b^2}{2} \Big[y \Big ]_0^h\\ A\amp = bh \amp Q_x \amp = \frac{h^2 b}{2} \amp Q_y \amp = \frac{b^2 h}{2} \end{align*}
3. Find the centroid.

   Substituting the results into the definitions gives

   \begin{align*} \bar{x} \amp = \frac{Q_y}{A} \amp \bar{y} \amp = \frac{Q_x}{A}\\ \amp = \frac{b^2h}{2} \bigg/ { bh} \amp \amp = \frac{h^2b}{2} \bigg/ { bh}\\ \amp = \frac{b}{2}\amp \amp = \frac{h}{2}\text{.} \end{align*}

This solution demonstrates solving integrals using square elements and double integrals. Set the slider on the diagram to \(dx\;dy\) to see a representative element.

1. Set up the integrals.

   Set the slider on the diagram to \(dx\;dy\) to see a representative element.

   The bounding functions \(x=0\text{,}\) \(x=a\text{,}\) \(y = 0\) and \(y = h\text{.}\)

   Instead of strips, the integrals will be evaluated using square elements with width \(dx\) and height \(dy\) located at \((x,y)\text{.}\)

   The area of the square element is the base times the height, so

   \begin{equation*} dA = dx\ dy = dy\ dx\text{.} \end{equation*}

   The centroid of the strip is located at its midpoint so, by inspection

   \begin{align*} \bar{x}_{\text{el}} \amp = x \\ \bar{y}_{\text{el}} \amp = y \end{align*}

   We will integrate twice, first with respect to \(y\) and then with respect to \(x\text{.}\) The limits on the first integral are \(y = 0\) to \(h\) and \(x = 0\) to \(b\) on the second. For a rectangle, both \(b\) and \(h\) are constants. In other situations, the upper or lower limits may be functions of \(x\) or \(y\text{.}\)

2. Solve the integrals.

   Substitute \(dA\text{,}\) \(\bar{x}_{\text{el}}\text{,}\) and \(\bar{y}_{\text{el}}\) into (7.7.2) and integrate the ‘inside’ integral, then the ‘outside’ integral. The results are the same as before.

   \begin{align*} A \amp = \int dA \amp Q_x \amp = \int \bar{y}_{\text{el}}\ dA \amp Q_y \amp = \int \bar{x}_{\text{el}}\ dA \\ \amp = \int_0^b\int_0^h dy\ dx \amp \amp = \int_0^b\int_0^h y\ dy\ dx \amp \amp = \int_0^b \int_0^h x\ dy\ dx\\ \amp = \int_0^b \left[ \int_0^h dy \right] dx \amp \amp = \int_0^b \left[\int_0^h y\ dy\right] dx \amp \amp = \int_0^b x \left[ \int_0^h dy\right] dx\\ \amp = \int_0^b \Big[ y \Big]_0^h dx \amp \amp = \int_0^b \Big[ \frac{y^2}{2} \Big]_0^h dx \amp \amp = \int_0^b x \Big[ y \Big]_0^h dx\\ \amp = h \int_0^b dx \amp \amp = \frac{h^2}{2} \int_0^b dx \amp \amp = h\int_0^b x\ dx\\ \amp = h\Big [ x \Big ]_0^b \amp \amp =\frac{h^2}{2} \Big [ x \Big ]_0^b \amp \amp = h \Big [ \frac{x^2}{2} \Big ]_0^b \\ A\amp = hb \amp Q_x\amp = \frac{h^2b}{2} \amp Q_y \amp = \frac{b^2 h}{2} \end{align*}
3. Find the centroid.

   Substituting the results into the definitions gives

   \begin{align*} \bar{x} \amp = \frac{Q_y}{A} \amp \bar{y} \amp = \frac{Q_x}{A}\\ \amp = \frac{b^2h}{2} \bigg/ { bh} \amp \amp = \frac{h^2b}{2} \bigg/ { bh}\\ \amp = \frac{b}{2}\amp \amp = \frac{h}{2}\text{.} \end{align*}

This solution demonstrates finding the centroid of the triangle using vertical strips \(dA = y\ dx\text{.}\) Set the slider on the diagram to \(y\;dx\) to see a representative element.

1. Set up the integrals.

   The bounding functions in this example are the \(x\) axis, the vertical line \(x = b\text{,}\) and the straight line through the origin with a slope of \(\frac{h}{b}\text{.}\) Using the slope-intercept form of the equation of a line, the upper bounding function is

   \begin{equation*} y = f(x) = \frac{h}{b} x \end{equation*}

   and any point on this line is designated \((x,y)\text{.}\)

   The strip extends from \((x,0)\) on the \(x\) axis to \((x,y)\) on the function, has a height of \(y\text{,}\) and a differential width \(dx\text{.}\) The area of this strip is

   \begin{equation*} dA = y dx\text{.} \end{equation*}

   The centroid of the strip is located at its midpoint so, by inspection

   \begin{align*} \bar{x}_{\text{el}} \amp = x \\ \bar{y}_{\text{el}} \amp = y/2 \end{align*}

   With vertical strips the variable of integration is \(x\text{,}\) and the limits are \(x=0\) to \(x=b\text{.}\)

2. Solve the integrals.

   Substitute \(dA\text{,}\) \(\bar{x}_{\text{el}}\text{,}\) and \(\bar{y}_{\text{el}}\) into (7.7.2) and integrate. In contrast to the rectangle example both \(dA\) and \(\bar{y}_{\text{el}}\) are functions of \(x\text{,}\) and will have to be integrated accordingly.

   \begin{align*} A \amp = \int dA \amp Q_x \amp = \int \bar{y}_{\text{el}}\ dA \amp Q_y \amp = \int \bar{x}_{\text{el}}\ dA \\ \amp = \int_0^b y\ dx \amp \amp = \int_0^b \frac{y}{2} (y\ dx ) \amp \amp = \int_0^b x\; (y\ dx)\\ \amp = \int_0^b \frac{h}{b}x\ dx \amp \amp = \frac{1}{2} \int_0^b \left(\frac{h}{b} x\right)^2\ dx \amp \amp = \int_0^b x\; \left(\frac{h}{b} x \right) \ dx\\ \amp = \frac{h}{b} \Big [ \frac{x^2}{2} \Big ]_0^b \amp \amp = \frac{h^2}{2 b^2} \int_0^b x^2 dx \amp \amp = \frac{h}{b} \int_0^b x^2 \ dx\\ \amp = \frac{h}{\cancel{b}} \frac{b^{\cancel{2}}}{2} \amp \amp = \frac{h^2}{2b^2} \Big [\frac{x^3}{3} \Big ]_0^b \amp \amp = \frac{h}{b} \left[\frac{x^3}{3} \right ]_0^b\\ A \amp =\frac{bh}{2} \amp Q_x \amp = \frac{h^2 b}{6} \amp Q_y \amp = \frac{b^2 h}{3} \end{align*}

   We learn that the area of a triangle is one half base times height. Since the area formula is well known, it would have been more efficient to skip the first integral. Note that \(A\) has units of \([\text{length}]^2\text{,}\) and \(Q_x\) and \(Q_y\) have units of \([\text{length}]^3\text{.}\)

3. Find the Centroid.

   Substituting the results into the definitions gives

   \begin{align*} \bar{x} \amp = \frac{Q_y}{A} \amp \bar{y} \amp = \frac{Q_x}{A}\\ \amp = \frac{b^2h}{3} \bigg/ \frac{bh}{2} \amp \amp = \frac{h^2b}{6} \bigg/ \frac{bh}{2}\\ \amp = \frac{2}{3}b\amp \amp = \frac{1}{3}h\text{.} \end{align*}

This solution demonstrates solving integrals using horizontal rectangular strips. Set the slider on the diagram to \((b-x)\;dy\) to see a representative element.

1. Set up the integrals.

   As before, the triangle is bounded by the \(x\) axis, the vertical line \(x = b\text{,}\) and the line

   \begin{equation*} y = f(x) = \frac{h}{b} x\text{.} \end{equation*}

   To integrate using horizontal strips, the function \(f(x)\) must be inverted to express \(x\) in terms of \(y\text{.}\) Solving for \(f(x)\) for \(x\) gives

   \begin{equation*} x = g(y) = \frac{b}{h} y\text{.} \end{equation*}

   The limits on the integral are from \(y = 0\) to \(y = h\text{.}\)

   The strip extends from \((x,y)\) to \((b,y)\text{,}\) has a height of \(dy\text{,}\) and a length of \((b-x)\text{,}\) therefore the area of this strip is

   \begin{equation*} dA = (b-x) dy\text{.} \end{equation*}

   The coordinates of the midpoint of the element are

   \begin{align*} \bar{y}_{\text{el}} \amp = y\\ \bar{x}_{\text{el}} \amp = x + \frac{(b-x)}{2} = \frac{b+x}{2}\text{.} \end{align*}

   These expressions are recognized as the average of the \(x\) and \(y\) coordinates of strip’s endpoints.

   A common student mistake is to use \(dA = x\ dy\text{,}\) and \(\bar{x}_{\text{el}} = x/2\text{.}\) These would be correct if you were looking for the properties of the area to the left of the curve.

2. Solve the integrals.

   Substitute \(dA\text{,}\) \(\bar{x}_{\text{el}}\text{,}\) and \(\bar{y}_{\text{el}}\) into the definitions of \(Q_x\) and \(Q_y\) and integrate. The results are the same as we found using vertical strips. There in no need to evaluate \(A = \int dA\) since we know that \(A = \frac{bh}{2}\) for a triangle.

   \begin{align*} Q_x \amp = \int \bar{y}_{\text{el}}\ dA \amp Q_y \amp = \int \bar{x}_{\text{el}}\ dA \\ \amp = \int_0^h y\ (b-x) \ dy \amp \amp = \int_0^h \frac{(b+x)}{2} (b-x)\ dy\\ \amp = \int_0^h \left( by - xy\right) \ dy \amp \amp = \frac{1}{2}\int_0^h \left(b^2-x^2\right)\ dy\\ \amp = \int_0^h \left( by -\frac{by^2}{h}\right) dy \amp \amp = \frac{1}{2}\int_0^h\left( b^2 - \frac{b^2y^2}{h^2}\right) dy\\ \amp = b \Big [\frac{ y^2}{2} - \frac{y^3}{3h} \Big ]_0^h \amp \amp = \frac{b^2}{2} \Big[y - \frac{y^3}{3 h^2}\Big ]_0^h\\ \amp = bh^2 \Big (\frac{1}{2} - \frac{1}{3} \Big ) \amp \amp = \frac{1}{2}( b^2h) \Big(1 - \frac{1}{3}\Big )\\ Q_x \amp = \frac{h^2 b}{6} \amp Q_y \amp = \frac{b^2 h}{3} \end{align*}

   It makes solving these integrals easier if you avoid prematurely substituting in the function for \(x\) and if you factor out constants whenever possible. Here it \(x = g(y)\) was not substituted until the fourth line.

3. Find the centroid.

   Substituting the results into the definitions gives

   \begin{align*} \bar{x} \amp = \frac{Q_y}{A} \amp \bar{y} \amp = \frac{Q_x}{A}\\ \amp = \frac{b^2h}{3} \bigg/ \frac{bh}{2} \amp \amp = \frac{h^2b}{6} \bigg/ \frac{bh}{2}\\ \amp = \frac{2}{3}b\amp \amp = \frac{1}{3}h\text{.} \end{align*}

This solution demonstrates solving integrals using square elements and double integrals. Set the slider on the diagram to \(dx\;dy\) or \(dy\;dx\) to see a representative element.

1. Set up the integrals.

   As before, the triangle is bounded by the \(x\) axis, the vertical line \(x = b\text{,}\) and the line

   \begin{equation*} y = f(x) = \frac{h}{b} x \quad \text{or in terms of } y, \quad x = g(y) = \frac{b}{h} y\text{.} \end{equation*}

   In this solution the integrals will be evaluated using square differential elements \(dA=dy\; dx\) located at \((x,y)\text{.}\)

   With double integration, you must take care to evaluate the limits correctly, since the limits on the inside integral are functions of the variable of integration of the outside integral. The inside integral essentially stacks the elements into strips and the outside integral adds all the strips to cover the area.

   Choosing to express \(dA\) as \(dy\;dx\) means that the integral over \(y\) will be conducted first. The limits on the inside integral are from \(y = 0\) to \(y = f(x)\text{.}\) Then, the limits on the outside integral are from \(x = 0\) to \(x=b.\)

   Using \(dA= dx\;dy\) would reverse the order of integration, so the inside integral’s limits would be from \(x = g(y)\) to \(x = b\text{,}\) and the limits on the outside integral would be \(y=0\) to \(y = h\text{.}\) Either choice will give the same results — if you don't make any errors!

   The area of the square element is the base times the height, so

   \begin{equation*} dA = dy\ dx\text{.} \end{equation*}

   The centroid of the square is located at its midpoint so, by inspection

   \begin{align*} \bar{x}_{\text{el}} \amp = x \\ \bar{y}_{\text{el}} \amp = y \end{align*}
2. Solve the integrals.

   Substitute \(dA\text{,}\) \(\bar{x}_{\text{el}}\text{,}\) and \(\bar{y}_{\text{el}}\) into (7.7.2) and integrate the ‘inside’ integral, then the ‘outside’ integral. The results are the same as before.

   \begin{align*} Q_x \amp = \int \bar{y}_{\text{el}}\ dA \amp Q_y \amp = \int \bar{x}_{\text{el}}\ dA \\ \amp = \int_0^b\int_0^{f(x)} y\ dy\ dx \amp \amp = \int_0^b \int_0^{f(x)} x\ dy\ dx\\ \amp = \int_0^b \left[\int_0^{f(x)} y\ dy\right] dx \amp \amp = \int_0^b x \left[ \int_0^{f(x)} dy\right] dx\\ \amp = \int_0^b \left[ \frac{y^2}{2} \right]_0^{f(x)} dx \amp \amp = \int_0^b x \bigg[ y \bigg]_0^{f(x)} dx\\ \amp = \frac{1}{2}\int_0^b \left[ \frac{h^2}{b^2} x^2 \right] dx \amp \amp = \int_0^b x \left[ \frac{h}{b} x \right] dx\\ \amp = \frac{h^2}{2b^2} \int_0^b x^2 dx \amp \amp = \frac{h}{b}\int_0^b x^2\ dx\\ \amp =\frac{h^2}{2b^2} \Big [\frac{x^3}{3} \Big ]_0^b \amp \amp = \frac{h}{b} \Big [ \frac{x^3}{3} \Big ]_0^b \\ Q_x \amp = \frac{h^2 b}{6} \amp Q_y \amp = \frac{b^2 h}{3} \end{align*}
3. Find the centroid.

   Substituting Q_x and \(Q_y\) along with \(A = bh/2\) into the centroid definitions gives

   \begin{align*} \bar{x} \amp = \frac{Q_y}{A} \amp \bar{y} \amp = \frac{Q_x}{A}\\ \amp = \frac{b^2h}{3} \bigg/ \frac{bh}{2} \amp \amp = \frac{h^2b}{6} \bigg/ \frac{bh}{2}\\ \amp = \frac{2}{3}b\amp \amp = \frac{1}{3}h\text{.} \end{align*}

We will use (7.7.2) with vertical strips to find the centroid of a spandrel.

1. Set up the integrals.

   Determining the bounding functions and setting up the integrals is usually the most difficult part of problems like this. Begin by drawing and labeling a sketch of the situation.

   1. Place a point in the first quadrant and label it \(P=(a,b)\text{.}\) This point is in the first quadrant and fixed since we are told that \(a\) and \(b\) are positive integers

   2. Place a horizontal line through \(P\) to make the upper bound.

   3. Sketch in a parabola with a vertex at the origin and passing through \(P\) and shade in the enclosed area.

   4. Decide which differential element you intend to use. For this example we choose to use vertical strips, which you can see if you tick show strips in the interactive above. Horizontal strips \(dA = x\ dy\) would give the same result, but you would need to define the equation for the parabola in terms of \(y\text{.}\)

   Determining the equation of the parabola and expressing it in terms of of \(x\) and any known constants is a critical step. You should remember from algebra that the general equation of parabola with a vertex at the origin is \(y = k x^2\text{,}\) where \(k\) is a constant which determines the shape of the parabola. If \(k \gt 0\text{,}\) the parabola opens upward and if \(k \lt 0\text{,}\) the parabola opens downward.

   To find the value of \(k\text{,}\) substitute the coordinates of \(P\) into the general equation, then solve for \(k\text{.}\)

   \begin{align*} y \amp = k x^2, \text{ so at } P \\ (b) \amp = k (a)^2\\ k \amp= \frac{b}{a^2} \end{align*}

   The resulting function of the parabola is

   \begin{equation*} y = y(x) = \frac{b}{a^2} x^2\text{.} \end{equation*}

   To perform the integrations, express the area and centroidal coordinates of the element in terms of the points at the top and bottom of the strip. The area of the strip is its height times its base, so

   \begin{equation*} dA = (b-y)\ dx\text{.} \end{equation*}

   If you incorrectly used \(dA = y\ dx\text{,}\) you would find the centroid of the spandrel below the curve.

   For vertical strips, the bottom is at \((x,y)\) on the parabola, and the top is directly above at \((x,b)\text{.}\) The strip has a differential width \(dx\text{.}\) The centroid of the strip is located at its midpoint and the coordinates are are found by averaging the \(x\) and \(y\) coordinates of the points at the top and bottom.

   \begin{align*} \bar{x}_{\text{el}} \amp = (x + x)/2 = x\\ \bar{y}_{\text{el}} \amp = (y+b)/2 \end{align*}

   For vertical strips, the integrations are with respect to \(x\text{,}\) and the limits on the integrals are \(x=0\) on the left to \(x = a\) on the right.

2. Solve the integrals.

   We have already established that \(y(x) = k x^2\) where \(k = b/a^2\text{.}\) To simplify the algebra, it is best not to prematurely substitute y(x) and \(k\text{,}\) but you must substitute in any functions of \(x\) before you do the integration step.

   \begin{align*} A \amp = \int dA \amp Q_x \amp = \int \bar{y}_{\text{el}} dA \amp Q_y \amp = \int \bar{x}_{\text{el}} dA \\ \amp = \int_0^a (b-y)\ dx \amp \amp = \int_0^a \frac{(b+y)}{2} (b-y) dx \amp \amp = \int_0^a x (b-y)\ dx\\ \amp = \int_0^a (b-kx^2)\ dx \amp \amp = \frac{1}{2}\int_0^a (b^2-y^2)\ dx \amp \amp = \int_o^a x (b-y) \ dx\\ \amp = \left . bx - k \frac{x^3}{3} \right |_0^a \amp \amp = \frac{1}{2} \int_0^a (b^2-(k x^2)^2)\ dx \amp \amp = \int_o^a x (b-k x^2) \ dx\\ \amp = ba - k \frac{a^3}{3} \amp \amp = \frac{1}{2} \int_0^a (b^2-k^2 x^4)\ dx \amp \amp = \int_o^a (bx-k x^3) \ dx\\ \amp = ba - \left(\frac{b}{a^2}\right)\frac{a^3}{3} \amp \amp = \frac{1}{2} \left[b^2 x - k^2 \frac{x^5}{5} \right ]_0^a \amp \amp = \left[\frac{bx^2}{2} - k \frac{x^4}{4}\right ]_0^a\\ \amp = \frac{3ba}{3} - \frac{ba}{3} \amp \amp = \frac{1}{2} \left[b^2 a - \left(\frac{b}{a^2}\right)^2 \frac{a^5}{5} \right ] \amp \amp = \left[\frac{ba^2}{2} - \left(\frac{b}{a^2}\right) \frac{4^4}{4}\right ]\\ \amp = \frac{2}{3} ba \amp \amp = \frac{1}{2} b^2a \left[1-\frac{1}{5}\right] \amp \amp = ba^2\left[\frac{1}{2} - \frac{1}{4}\right]\\ A \amp = \frac{2}{3} ba \amp Q_x \amp = \frac{2}{5} b^2a \amp Q_y \amp = \frac{1}{4} ba^2 \end{align*}

   The area of the spandrel is \(2/3\) of the area of the enclosing rectangle and the moments of area have units of \([\text{length}]^3\text{.}\)

3. Find the centroid.

   Substituting the results into the definitions gives

   \begin{align*} \bar{x} \amp = \frac{Q_y}{A} \amp \bar{y} \amp = {Q_x}{A}\\ \amp = \frac{ba^2}{4 } \bigg/ \frac{2 ba}{3} \amp \amp = \frac{2 b^2a }{5}\bigg/ \frac{2 ba}{3}\\ \amp = \frac{3}{8} a \amp \amp = \frac{2}{5} b\text{.} \end{align*}

   \(\bar{x}\) is \(3/8\) of the width and \(\bar{y}\) is \(2/5\) of the height of the enclosing rectangle.

This solution demonstrates finding the centroid of the area between two functions using vertical strips \(dA = y\ dx\text{.}\) Set the slider on the diagram to \(h\;dx\) to see a representative element.

1. Set up the integrals.

   The bounding functions in this example are the \(x\) axis, the vertical line \(x = b\text{,}\) and the straight line through the origin with a slope of \(\frac{h}{b}\text{.}\) Using the slope-intercept form of the equation of a line, the upper bounding function is

   \begin{equation*} y = f(x) = \frac{h}{b} x \end{equation*}

   and any point on this line is designated \((x,y)\text{.}\)

   The strip extends from \((x,0)\) on the \(x\) axis to \((x,y)\) on the function, has a height of \(y\text{,}\) and a differential width \(dx\text{.}\) The area of this strip is

   \begin{equation*} dA = y dx\text{.} \end{equation*}

   The centroid of the strip is located at its midpoint so, by inspection

   \begin{align*} \bar{x}_{\text{el}} \amp = x \\ \bar{y}_{\text{el}} \amp = y/2 \end{align*}

   With vertical strips the variable of integration is \(x\text{,}\) and the limits are \(x=0\) to \(x=b\text{.}\)

2. Solve the integrals.

   Substitute \(dA\text{,}\) \(\bar{x}_{\text{el}}\text{,}\) and \(\bar{y}_{\text{el}}\) into (7.7.2) and integrate. In contrast to the rectangle example both \(dA\) and \(\bar{y}_{\text{el}}\) are functions of \(x\text{,}\) and will have to be integrated accordingly.

   \begin{align*} A \amp = \int dA \\ \amp = \int_0^{1/2} (y_1 - y_2) \ dx \\ \amp = \int_0^{1/2} \left (\frac{x}{4} - \frac{x^2}{2}\right) \ dx \\ \amp = \Big [ \frac{x^2}{8} - \frac{x^3}{6} \Big ]_0^{1/2} \\ \amp = \Big [ \frac{1}{32} - \frac{1}{48} \Big ] \\ A \amp =\frac{1}{96} \end{align*}
   \begin{align*} Q_x \amp = \int \bar{y}_{\text{el}}\ dA \amp Q_y \amp = \int \bar{x}_{\text{el}}\ dA \\ \amp = \int_0^{1/2} \left(\frac{y_1+y_2}{2} \right) (y_1-y_2)\ dx \amp \amp = \int_0^{1/2} x(y_1-y_2)\ dx\\ \amp = \frac{1}{2} \int_0^{1/2} \left(y_1^2 - y_2^2 \right)\ dx \amp \amp = \int_0^{1/2} x\left(\frac{x}{4} - \frac{x^2}{2}\right) \ dx\\ \amp = \frac{1}{2} \int_0^{1/2} \left(\frac{x^2}{16} - \frac{x^4}{4}\right)\ dx\amp \amp = \int_0^{1/2}\left(\frac{x^2}{4} - \frac{x^3}{2}\right)\ dx\\ \amp = \frac{1}{2} \Big [\frac{x^3}{48}-\frac{x^5}{20} \Big ]_0^{1/2} \amp \amp = \left[\frac{x^3}{12}- \frac{x^4}{8} \right ]_0^{1/2}\\ \amp = \frac{1}{2} \Big [\frac{1}{384}-\frac{1}{640} \Big ] \amp \amp = \Big [\frac{1}{96}-\frac{1}{128} \Big ] \\ Q_x \amp = \frac{1}{1920} \amp Q_y \amp = \frac{1}{384} \end{align*}
3. Find the Centroid.

   Substituting the results into the definitions gives

   \begin{align*} \bar{x} \amp = \frac{Q_y}{A} \amp \bar{y} \amp = \frac{Q_x}{A}\\ \amp = \frac{1}{384} \bigg/ \frac{1}{96} \amp \amp = \frac{1}{1920} \bigg/ \frac{1}{96}\\ \bar{x} \amp= \frac{1}{4} \amp \bar{y}\amp =\frac{1}{20}\text{.} \end{align*}

This solution demonstrates finding the centroid of the area between two functions using vertical strips \(dA = y\ dx\text{.}\) Set the slider on the diagram to \(h\;dx\) to see a representative element.

1. Set up the integrals.

   The bounding functions in this example are the \(x\) axis, the vertical line \(x = b\text{,}\) and the straight line through the origin with a slope of \(\frac{h}{b}\text{.}\) Using the slope-intercept form of the equation of a line, the upper bounding function is

   \begin{equation*} y = f(x) = \frac{h}{b} x \end{equation*}

   and any point on this line is designated \((x,y)\text{.}\)

   The strip extends from \((x,0)\) on the \(x\) axis to \((x,y)\) on the function, has a height of \(y\text{,}\) and a differential width \(dx\text{.}\) The area of this strip is

   \begin{equation*} dA = y dx\text{.} \end{equation*}

   The centroid of the strip is located at its midpoint so, by inspection

   \begin{align*} \bar{x}_{\text{el}} \amp = x \\ \bar{y}_{\text{el}} \amp = y/2 \end{align*}

   With vertical strips the variable of integration is \(x\text{,}\) and the limits are \(x=0\) to \(x=b\text{.}\)

2. Solve the integrals.

   Substitute \(dA\text{,}\) \(\bar{x}_{\text{el}}\text{,}\) and \(\bar{y}_{\text{el}}\) into (7.7.2) and integrate. In contrast to the rectangle example both \(dA\) and \(\bar{y}_{\text{el}}\) are functions of \(x\text{,}\) and will have to be integrated accordingly.

   \begin{align*} A \amp = \int dA \\ \amp = \int_0^y (x_2 - x_1) \ dy \\ \amp = \int_0^{1/8} \left (4y - \sqrt{2y} \right) \ dy \\ \amp = \Big [ 2y^2 - \frac{4}{3} y^{3/2} \Big ]_0^{1/8} \\ \amp = \Big [ \frac{1}{32} - \frac{1}{48} \Big ] \\ A \amp =\frac{1}{96} \end{align*}
   \begin{align*} Q_x \amp = \int \bar{y}_{\text{el}}\ dA \amp Q_y \amp = \int \bar{x}_{\text{el}}\ dA \\ \amp = \int_0^{1/8} y (x_2-x_1)\ dy \amp \amp = \int_0^{1/8} \left(\frac{x_2+x_1}{2} \right) (x_2-x_1)\ dy\\ \amp = \int_0^{1/8} y \left(\sqrt{2y}-4y\right)\ dy \amp \amp = \frac{1}{2} \int_0^{1/8} \left(x_2^2 - x_1^2\right) \ dy\\ \amp = \int_0^{1/8} \left(\sqrt{2} y^{3/2} - 4y^2 \right)\ dy\amp \amp = \frac{1}{2} \int_0^{1/8}\left(2y -16 y^2\right)\ dy\\ \amp = \Big [\frac{2\sqrt{2}}{5} y^{5/2} -\frac{4}{3} y^3 \Big ]_0^{1/8} \amp \amp = \frac{1}{2} \left[y^2- \frac{16}{3}y^3 \right ]_0^{1/8}\\ \amp = \Big [\frac{1}{320}-\frac{1}{384} \Big ] \amp \amp = \frac{1}{2} \Big [\frac{1}{64}-\frac{1}{96} \Big ] \\ Q_x \amp = \frac{1}{1920} \amp Q_y \amp = \frac{1}{384} \end{align*}
3. Find the Centroid.

   Substituting the results into the definitions gives

   \begin{align*} \bar{x} \amp = \frac{Q_y}{A} \amp \bar{y} \amp = \frac{Q_x}{A}\\ \amp = \frac{1}{384} \bigg/ \frac{1}{96} \amp \amp = \frac{1}{1920} \bigg/ \frac{1}{96}\\ \bar{x} \amp= \frac{1}{4} \amp \bar{y}\amp =\frac{1}{20}\text{.} \end{align*}

We will use (7.7.2) with polar coordinates \((\rho, \theta)\) to solve this problem because they are a natural fit for the geometry. In polar coordinates, the equation for the bounding semicircle is simply

Normally this involves evaluating three integrals but as you will see, we can take some shortcuts in this problem. Otherwise we will follow the same procedure as before.

1. Set up the integrals.

   Divide the semi-circle into "rectangular" differential elements of area \(dA\text{,}\) as shown in the interactive when you select Show element. This shape is not really a rectangle, but in the limit as \(d\rho\) and \(d\theta\) approach zero, it doesn't make any difference.

   The radial height of the rectangle is \(d\rho\) and the tangential width is the arc length \(\rho d\theta\text{.}\) The product is the differential area \(dA\text{.}\)

   \begin{equation} dA = (d\rho)(\rho\ d\theta) = \rho\ d\rho\ d\theta\text{.}\tag{7.7.7} \end{equation}

   The differential element is located at \((\rho, \theta)\) in polar coordinates. Expressing this point in rectangular coordinates gives

   \begin{align*} \bar{x}_{\text{el}} \amp = \rho \cos \theta\\ \bar{y}_{\text{el}} \amp = \rho \sin \theta\text{.} \end{align*}
2. Solve the integrals.

   The area of a semicircle is well known, so there is no need to actually evaluate \(A = \int dA\text{,}\)

   \begin{equation*} A = \int dA = \frac{\pi r^2}{2}\text{.} \end{equation*}

   Since the semi-circle is symmetrical about the \(y\) axis,

   \begin{equation*} Q_y = \int \bar{x}_{\text{el}}\; dA= 0\text{.} \end{equation*}

   This is because each element of area to the right of the \(y\) axis is balanced by a corresponding element the same distance the left which cancel each other out in the sum.

   All that remains is to evaluate the integral \(Q_x\) in the numerator of

   \begin{equation*} \bar{y} = \frac{Q_x}{A} = \frac{\bar{y}_{\text{el}}\; dA}{A} \end{equation*}

   The differential area \(dA\) is the product of two differential quantities, we will need to perform a double integration.

   \begin{align*} Q_x \amp = \int \bar{y}_{\text{el}} dA \\ \amp = \int_0^\pi \int_0^r (\rho \sin \theta) \rho \; d\rho\; d\theta\\ \amp = \int_0^\pi \sin \theta \left[ \int_0^r \rho^2 \; d\rho\right ] d\theta\\ \amp = \int_0^\pi \sin \theta \left[ \frac{\rho^3} {3}\right ]_0^r \; d\theta\\ \amp = \frac{r^3}{3} \ \int_0^\pi \sin \theta \; d\theta\\ \amp = \frac{r^3}{3} \left[ - \cos \theta \right]_0^\pi\\ \amp = -\frac{r^3}{3} \left[ \cos \pi - \cos 0 \right ]\\ \amp = -\frac{r^3}{3} \left[ (-1) - (1) \right ]\\ Q_x \amp = \frac{2}{3} r^3 \end{align*}
3. Find the centroid.

   Substituting the results into the definitions gives

   \begin{align*} \bar{y} \amp = \frac{Q_x}{A} \\ \amp = \frac{2 r^3}{3} \bigg/ \frac{\pi r^2}{2}\\ \amp = \frac{4r}{3\pi}\text{.} \end{align*}

   So \(\bar{x}=0\) and lies on the axis of symmetry, and \(\bar{y} =\dfrac{4r}{3\pi}\) above the diameter.

   This result can be extended by noting that a semi-circle is mirrored quarter-circles on either side of the \(y\) axis. These must have the same \(\bar{y}\) value as the semi-circle. Further, quarter-circles are symmetric about a \(\ang{45}\) line, so for the quarter-circle in the first quadrant,

   \begin{equation*} \bar{x} = \bar{y} = \frac{4r}{3\pi}\text{.} \end{equation*}