# Applications of Integration: Area and Volume

This section covers:

One very useful application of Integration is finding the area and volume of “curved” figures, that we couldn’t typically get without using Calculus. Since we already know that can use the integral to get the area between the $$x$$- and $$y$$-axis and a function, we can also get the volume of this figure by rotating the figure around either one of the axes.

# Area Between Curves

Since we know how to get the area under a curve here in the Definite Integrals section, we can also get the area between two curves by subtracting the bottom curve from the top curve everywhere where the top curve is higher than the bottom curve. The cool thing about this is it even works if one of the curves is below the $$x$$-axis, as long as the higher curve always stays above the lower curve in the integration interval.

Note that we may need to find out where the two curves intersect (and where they intersect the $$x$$-axis) to get the limits of integration. And sometimes we have to divide up the integral if the functions cross over each other in the integration interval.

Here is the formal definition of the area between two curves:

Area of Region Between Two Curves For functions $$f$$ and $$g$$ where $$f\left( x \right)\ge g\left( x \right)$$ for all $$x$$ in $$[a,b]$$, the area of the region bounded by the graphs and the vertical lines $$x=a$$ and $$x=b$$ is:

$$\text{Area}=\int\limits_{a}^{b}{{\left[ {f\left( x \right)-g\left( x \right)} \right]}}\,dx$$

Let’s try some problems:

 Integration Area Problem and Solution Set up the definite integral that gives the following area (don’t solve): $$\begin{array}{l}f\left( x \right)={{x}^{2}}-2x\\g\left( x \right)=0\end{array}$$ Solution: $$\int\limits_{0}^{2}{{\left[ {0-\left( {{{x}^{2}}-2x} \right)} \right]dx}}=-\int\limits_{0}^{2}{{\left( {{{x}^{2}}-2x} \right)dx}}$$ Set up and solve the definite integral that gives the following area: $$\begin{array}{l}f\left( x \right)={{x}^{2}}-5x+6\\g\left( x \right)=-{{x}^{2}}+x+6\end{array}$$ Solution:   \displaystyle \begin{align}&\int\limits_{0}^{3}{{\left[ {\left( {-{{x}^{2}}+x+6} \right)-\left( {{{x}^{2}}-5x+6} \right)} \right]dx}}\\\,\,\,&\,\,\,=\int\limits_{0}^{3}{{\left( {-2{{x}^{2}}+6x} \right)dx}}=\left[ {-\frac{2}{3}{{x}^{3}}+3{{x}^{2}}} \right]_{0}^{3}\\\,\,\,&\,\,\,=\left( {-\frac{2}{3}{{{\left( 3 \right)}}^{3}}+3{{{\left( 3 \right)}}^{2}}} \right)-\left( {-\frac{2}{3}{{{\left( 0 \right)}}^{3}}+3{{{\left( 0 \right)}}^{2}}} \right)=9\end{align} Set up and solve the definite integral that gives the following area (don’t solve): $$\begin{array}{l}f\left( \theta \right)=-\sin \theta \\g\left( \theta \right)=0\end{array}$$ Solution: Divide graph into two separate integrals, since from $$-\pi$$ to 0, $$f\left( \theta \right)\ge g\left( \theta \right)$$, and from 0 to $$\pi$$, $$g\left( \theta \right)\ge f\left( \theta \right)$$: \displaystyle \begin{align}&\int\limits_{{-\pi }}^{0}{{\left( {-\sin \theta -0} \right)d\theta }}+\int\limits_{0}^{\pi }{{\left[ {0-\left( {-\sin \theta } \right)} \right]d\theta }}\\&\,\,=\int\limits_{{-\pi }}^{0}{{\left( {-\sin \theta } \right)d\theta }}+\int\limits_{0}^{\pi }{{\left( {\sin \theta } \right)d\theta }}\\&\,\,=\left[ {\cos x} \right]_{{-\pi }}^{0}+\left[ {-\cos x} \right]_{0}^{\pi }\\&\,\,=\cos \left( 0 \right)-\cos \left( {-\pi } \right)+\left[ {-\cos \left( \pi \right)+\cos \left( 0 \right)} \right]\,\,\\&\,\,=1-\left( {-1} \right)+\left( {1+1} \right)=4\end{align} Sketch the region bounded by the graphs, and find the area: $$\displaystyle f\left( x \right)=\sqrt{x}+1,\,\,\,g\left( x \right)=\frac{1}{2}x+1$$ Solution: Draw the curves and set them equal to each other to see where the limits of integration will be: $$\displaystyle \sqrt{x}+1=\frac{1}{2}x+1;\,\,\,\,\sqrt{x}=\frac{1}{2}x;\,\,\,\,x=\frac{{{{x}^{2}}}}{4};\,\,\,\,4x={{x}^{2}}$$ $$\displaystyle {{x}^{2}}-4x=0;\,\,\,\,x\left( {x-4} \right)=0;\,\,\,x=0,\,\,4$$ Integrate from 0 to 4: \displaystyle \begin{align}&\int\limits_{0}^{4}{{\left[ {\left( {\sqrt{x}+1} \right)-\left( {\frac{1}{2}x+1} \right)} \right]dx}}=\int\limits_{0}^{4}{{\left( {{{x}^{{\frac{1}{2}}}}-\frac{x}{2}} \right)\,dx}}\\&\,\,\,=\left[ {\frac{2}{3}{{x}^{{\frac{3}{2}}}}-\frac{1}{4}{{x}^{2}}} \right]_{0}^{4}=\left[ {\frac{2}{3}{{{\left( 4 \right)}}^{{\frac{3}{2}}}}-\frac{1}{4}{{{\left( 4 \right)}}^{2}}} \right]-0=\frac{4}{3}\end{align}

Notice this next problem, where it’s much easier to find the area with respect to $$y$$, since we don’t have to divide up the graph. When we integrate with respect to $$y$$, we will have horizontal rectangles (parallel to the $$x$$-axis) instead of vertical rectangles (perpendicular to the $$x$$-axis), since we’ll use “$$dy$$” instead of “$$dx$$”. If we have the functions in terms of $$x$$, we need to use Inverse Functions to get them in terms of $$y$$.

 Integration Area Problem Rotated Around $$\boldsymbol {x}$$-axis Integration Area Problem Rotated Around $$\boldsymbol {y}$$-axis Sketch the region bounded by the graphs and find the area, with respect to $$x$$:   $$y=2x,\,\,\,y=2-2x,\,\,\,y=0$$   Solution: Draw the three lines and set equations equal to each other to get the limits of integration.   We need to divide the graph into two separate integrals, since the function “on top” changes from $$2x$$ to $$2-2x$$ at $$x=.5$$. (We can also get the intersection by setting the equations equal to each other:). We see $$x$$-intercepts are 0 and 1. The two separate integrals are from the intervals 0 to .5, and .5 to 1. (This area, a triangle, is $$\displaystyle \frac{1}{2}bh=\frac{1}{2}\cdot 1\cdot 1=.5$$.)   Integrating, we get this same area: \begin{align}&\int\limits_{0}^{{.5}}{{\left( {2x-0} \right)dx}}+\int\limits_{{.5}}^{1}{{\left[ {\left( {2-2x} \right)-0} \right]dx}}\\\,&\,\,=\int\limits_{0}^{{.5}}{{2x\,dx}}+\int\limits_{{.5}}^{1}{{\left( {2-2x} \right)dx}}\\\,&\,\,=\left. {{{x}^{2}}} \right|_{0}^{{.5}}+\left[ {2x-{{x}^{2}}} \right]_{{.5}}^{1}\\\,&\,\,={{\left( {.5} \right)}^{2}}-0+\left( {2\left( 1 \right)-{{{\left( 1 \right)}}^{2}}} \right)-\left( {2\left( {.5} \right)-{{{\left( {.5} \right)}}^{2}}} \right)\\\,&\,\,=.5\end{align} Sketch the region bounded by the graphs and find the area, with respect to $$y$$: $$y=2x,\,\,\,y=2-2x,\,\,\,y=0$$ Solution: If we use horizontal rectangles, we need to take the inverse of the functions to get $$x$$ in terms of $$y$$, so we have $$\displaystyle x=\frac{y}{2}$$ and $$\displaystyle x=\frac{{2-y}}{2}$$. We’ll integrate up the $$y$$-axis, from 0 to 1. Now we have one integral instead of two! Note the $$y$$ interval is from down to up, and the subtraction of functions is from right to left. \begin{align}&\int\limits_{0}^{1}{{\left( {\frac{{2-y}}{2}-\frac{y}{2}} \right)dy}}=\frac{1}{2}\int\limits_{0}^{1}{{\left( {2-2y} \right)dy}}\\&\,\,=\frac{1}{2}\left[ {2y-{{y}^{2}}} \right]_{0}^{1}=\frac{1}{2}\left( {1-0} \right)=.5\end{align} Note that some find it easier to think about rotating the graph 90° clockwise, which will yield its inverse. Then integrate with respect to $$x$$: \begin{align}&\int\limits_{0}^{1}{{\left( {\frac{{2-x}}{2}-\frac{x}{2}} \right)dx}}=\frac{1}{2}\int\limits_{0}^{1}{{\left( {2-2x} \right)dx}}\\&\,\,=\frac{1}{2}\left[ {2x-{{x}^{2}}} \right]_{0}^{1}=\frac{1}{2}\left( {1-0} \right)=.5\end{align}

Here are more problems where we take the area with respect to $$y$$:

 Integration Area Problem Integration Area Problem Sketch the region bounded by the graphs and find the area, with respect to $$y$$: $$f\left( y \right)=y\left( {4-y} \right),\,\,\,\,g\left( y \right)=-y$$   Solution:  Find points of intersection: $$\begin{array}{c}y\left( {4-y} \right)=-y;\,\,\,\,4y-{{y}^{2}}+y=0;\,\,\,\\y\left( {5-y} \right)=0;\,\,\,y=0,\,5\end{array}$$ The points of intersection are $$(-5,5)$$ and $$(0,0)$$.   Now graph. If you’re not sure how to graph, you can always make $$t$$-charts. Remember we go down to up for the interval, and right to left for the subtraction of functions: Now integrate: \begin{align}&\int\limits_{0}^{5}{{\left[ {\left( {4y-{{y}^{2}}} \right)-\left( {-y} \right)} \right]dy}}=\int\limits_{0}^{5}{{\left( {5y-{{y}^{2}}} \right)dy}}\\\,&\,\,=\left[ {\frac{5}{2}{{y}^{2}}-\frac{1}{3}{{y}^{3}}} \right]_{0}^{5}=\left( {\frac{5}{2}{{{\left( 5 \right)}}^{2}}-\frac{1}{3}{{{\left( 5 \right)}}^{3}}} \right)-0\\&\,\,=\frac{{125}}{6}\end{align} Sketch the region bounded by the graphs and find the area, with respect to $$y$$: $$f\left( y \right)={{y}^{2}}+2,\,\,\,g\left( y \right)=0,\,\,\,y=-1,\,\,\,y=2$$   Solution:  Graph first to verify the points of intersection. If you’re not sure how to graph, you can always make $$t$$-charts. Remember we go down to up for the interval, and right to left for the subtraction of functions:   We can see that we’ll use $$y=-1$$ and $$y=2$$ for the limits of integration: Now integrate: \begin{align}&\int\limits_{{-1}}^{2}{{\left[ {\left( {{{y}^{2}}+2} \right)-\left( 0 \right)} \right]dy}}=\int\limits_{{-1}}^{2}{{\left( {{{y}^{2}}+2} \right)dy}}\\&\,\,=\left[ {\frac{1}{3}{{y}^{3}}+2y} \right]_{{-1}}^{2}=\left( {\frac{1}{3}{{{\left( 2 \right)}}^{3}}+2\left( 2 \right)} \right)-\left( {\frac{1}{3}{{{\left( {-1} \right)}}^{3}}+2\left( {-1} \right)} \right)\\&\,\,=9\end{align}

# Volumes of Solids by Cross Sections

Now that we know how to get areas under and between curves, we can use this method to get the volume of a three-dimensional solid, either with cross sections, or by rotating a curve around a given axis. Think about it; every day engineers are busy at work trying to figure out how much material they’ll need for certain pieces of metal, for example, and they are using calculus to figure this stuff out!

Let’s first talk about getting the volume of solids by cross-sections of certain shapes. When doing these problems, think of the bottom of the solid being flat on your horizontal paper, and the 3-D part of it coming up from the paper. Cross sections might be squares, rectangles, triangles, semi-circles, trapezoids, or other shapes. We’ll have to use some geometry to get these areas.

Cross sections can either be perpendicular to the $$x$$-axis or $$y$$-axis; in our examples, they will be perpendicular to the $$x$$-axis, which is what is we are used to.

Given the cross sectional area $$A(x)$$ in interval [$$[a,b]$$, and cross sections are perpendicular to the  $$x$$-axis, the volume of this solid is $$\text{Volume = }\int\limits_{a}^{b}{{A\left( x \right)}}\,dx$$

Here are examples of volumes of cross sections between curves. Slices of the volume are shown to better see how the volume is obtained:

 Volumes of Solids by Cross Sections Squares   Set up the integral to find the volume of solid whose base is bounded by the graph of $$f\left( x \right)=\sqrt{{\sin \left( x \right)}}$$,  $$x=0,\,x=\pi$$, and the $$x$$-axis, with perpendicular cross sections that are squares. Note that the side of the square is the distance between the function and $$x$$-axis ($$b$$), and the area is $${{b}^{2}}$$. $$\displaystyle \text{Volume}=\int\limits_{0}^{\pi }{{{{{\left[ {\sqrt{{\sin \left( x \right)}}-0} \right]}}^{2}}\,dx}}=\int\limits_{0}^{\pi }{{\sin \left( x \right)}}\,dx$$ Semicircles   Set up the integral to find the volume of solid whose base is bounded by graphs of $$y=4x$$ and $$y={{x}^{2}}$$, with perpendicular cross sections that are semicircles. Note that the diameter ($$2r$$) of the semicircle is the distance between the curves, so the radius $$r$$ of each semicircle is $$\displaystyle \frac{{4x-{{x}^{2}}}}{2}$$. Thus, the area of each semicircle is $$\displaystyle \frac{{\pi {{r}^{2}}}}{2}=\frac{1}{2}\pi \cdot {{\left( {\frac{{4x-{{x}^{2}}}}{2}} \right)}^{2}}$$. Thus: $$\displaystyle \text{Volume}=\frac{1}{2}\pi \int\limits_{0}^{4}{{{{{\left[ {\frac{{\left( {4x-{{x}^{2}}} \right)}}{2}} \right]}}^{2}}}}dx=\frac{\pi }{8}\int\limits_{0}^{4}{{{{{\left( {4x-{{x}^{2}}} \right)}}^{2}}}}\,dx$$ Equilateral Triangles   Set up the integral to find the volume of solid whose base is bounded by the circle $${{x}^{2}}+{{y}^{2}}=9$$, with perpendicular cross sections that are equilateral triangles. (Area of equilateral triangle with side $$s$$ is $${{b}^{2}}$$.) Note that one of the sides of the triangle is twice the $$y$$ value of the function $$y=\sqrt{{9-{{x}^{2}}}}$$, and area is $$\displaystyle \frac{{\sqrt{3}}}{4}{{s}^{2}}=\frac{{\sqrt{3}}}{4}{{\left( {2\sqrt{{9-{{x}^{2}}}}} \right)}^{2}}$$. Thus: $$\displaystyle \text{Volume}=\frac{{\sqrt{3}}}{4}\int\limits_{{-3}}^{3}{{{{{\left( {2\sqrt{{9-{{x}^{2}}}}} \right)}}^{2}}}}dx=\sqrt{3}\int\limits_{{-3}}^{3}{{\left( {9-{{x}^{2}}} \right)}}\,dx$$ Rectangles   Set up to find the volume of solid whose base is bounded by the graphs of  $$y=.25{{x}^{2}}$$ and $$y=1$$, with perpendicular cross sections that are rectangles with height twice the base. Note that the base of the rectangle is $$1-.25{{x}^{2}}$$, the height of the rectangle is $$2\left( {1-.25{{x}^{2}}} \right)$$, and area is $$\text{base}\cdot \text{height}$$: \displaystyle \begin{align}\text{Volume}&=\int\limits_{{-2}}^{2}{{\left[ {\left( {1-.25{{x}^{2}}} \right)\cdot 2\left( {1-.25{{x}^{2}}} \right)} \right]dx}}\\&=2\int\limits_{{-2}}^{2}{{{{{\left( {1-.25{{x}^{2}}} \right)}}^{2}}}}\,dx\end{align}

Here’s one more that’s a little tricky:

 Volumes of Solids by Cross Sections Find the volume of a solid whose base is bounded by $$y={{x}^{3}},\,x=2$$, and the $$x$$-axis, and whose cross sections are perpendicular to the $$y$$-axis and are isosceles right triangles with a leg on the base of the solid. This one’s tricky since the cross sections are perpendicular to the $$y$$-axis which means we need to get the area with respect to $$y$$ and not $$x$$. Since we are given $$y$$ in terms of $$x$$, we’ll take the inverse of $$y={{x}^{3}}$$ to get $$x=\sqrt{y}$$.   First graph and find the points of intersection. When we get the area with respect to $$y$$, we use smaller to larger for the interval, and right to left to subtract the functions. Thus, we can see that each base, $$b$$, will be $$2-\sqrt{y}$$. The area of an isosceles triangle is $$\displaystyle A=\frac{1}{2}bh=\frac{1}{2}{{b}^{2}}$$, so our integral is: $$\displaystyle \text{Volume}=\int\limits_{{y=0}}^{{y=8}}{{\frac{1}{2}{{{\left( {2-\sqrt{y}} \right)}}^{2}}dy}}\approx 1.6$$.

# Volumes of Solids: The Disk Method

Now let’s talk about getting a volume by revolving a function or curve around a given axis to obtain a solid of revolution.

Since we know now how to get the area of a region using integration, we can get the volume of a solid by rotating the area around a line, which results in a right cylinder, or disk. (Remember that the formula for the volume of a cylinder is $$\pi {{r}^{2}}\cdot \text{height}$$). Note that the radius is the distance from the axis of revolution to the function, and the “height” of each disk, or slice is “$$dx$$”:

 Volumes of Solids: The Disk Method Horizontal Axis of Revolution:         $$\text{Volume}=\pi \int\limits_{a}^{b}{{{{{\left[ {f\left( x \right)} \right]}}^{2}}}}\,dx$$ Vertical Axis of Revolution: $$\text{Volume}=\pi \int\limits_{a}^{b}{{{{{\left[ {f\left( y \right)} \right]}}^{2}}}}\,dy$$ Let’s try some problems:

 Volume Using Disk Method Problem Volume Using Disk Method Problem Set up and solve the integral that gives the volume of the solid formed by rotating the region about the $$x$$-axis: $$y=0,\,\,\,y=16-{{x}^{2}}$$   Solution: Find where the functions intersect: $$\displaystyle 16-{{x}^{2}}=0;\,\,\,x=\pm 4$$ Now graph. If you’re not sure how to graph, you can always make t-charts. Notice that the radius of each circle will be the $$y$$ part of the function, or $$16-{{x}^{2}}$$. The formula for the volume is $$\pi \,\int\limits_{a}^{b}{{{{{\left[ {f\left( x \right)} \right]}}^{2}}}}\,dx$$. Now integrate: \begin{align}&\pi \int\limits_{{-4}}^{4}{{\left( {16-{{x}^{2}}} \right)dx}}\\&\,=\pi \left[ {16x-\frac{1}{3}{{x}^{3}}} \right]_{{-4}}^{4}\\\,&=\pi \left( {\left[ {16\left( 4 \right)-\frac{1}{3}{{{\left( 4 \right)}}^{3}}} \right]-\left[ {16\left( {-4} \right)-\frac{1}{3}{{{\left( {-4} \right)}}^{3}}} \right]} \right)\\&=\frac{{256}}{3}\pi \end{align} Set up (don’t solve) the integral that gives the volume of the solid formed by rotating the region bounded by the equations $$y=2\sqrt{x},\,\,\,y=0,\,\,\,x=9$$ about the given lines:  (a) Rotated around the $$x$$-axis (b) Rotated around the line $$x=9$$   Solution: Draw first: (a) (b) (a) Since the rotation is around the $$x$$-axis, the radius of each circle will be the $$x$$-axis part of the function, or $$2\sqrt{x}$$. The function hits the $$x$$-axis at 0 and 9, so the volume is $$\displaystyle \pi \int\limits_{0}^{9}{{{{{\left( {2\sqrt{x}} \right)}}^{2}}dx}}=2\pi \int\limits_{0}^{9}{{4x\,dx}}$$.   (b) This one’s tricky. First, to get $$y$$ in terms of $$x$$, we solve for the inverse of $$y=2\sqrt{x}$$ to get $$\displaystyle x={{\left( {\frac{y}{2}} \right)}^{2}}=\frac{{{{y}^{2}}}}{4}$$ (think of the whole graph being tilted sideways, and switching the $$x$$ and $$y$$ axes). Since we are rotating around the line $$x=9$$, to get a radius for the shaded area, we need to use $$\displaystyle 9-\frac{{{{y}^{2}}}}{4}$$ instead of just $$\displaystyle \frac{{{{y}^{2}}}}{4}$$ for the radius of the circles of the shaded region (try with real numbers and you’ll see). Thus, the volume is $$\displaystyle \pi \int\limits_{0}^{6}{{{{{\left( {9-\frac{{{{y}^{2}}}}{4}} \right)}}^{2}}dy}}$$.

# Volumes of Solids: The Washer Method

The washer method is similar to the disk method, but it covers solids of revolution that have “holes”, where we have inner and outer functions, thus inner and outer radii.

So now we have two revolving solids and we basically subtract the area of the inner solid from the area of the outer one. Note that for this to work, the middle function must be completely inside (or touching) the outer function over the integration interval.

 Volumes of Solids: The Washer Method Horizontal Axis of Revolution: $$\text{Volume}=\pi \int\limits_{a}^{b}{{\left( {{{{\left[ {R\left( x \right)} \right]}}^{2}}-{{{\left[ {r\left( x \right)} \right]}}^{2}}} \right)}}\,dx$$ Vertical Axis of Revolution: $$\text{Volume}=\pi \,\int\limits_{a}^{b}{{\left( {{{{\left[ {R\left( y \right)} \right]}}^{2}}-{{{\left[ {r\left( y \right)} \right]}}^{2}}} \right)}}\,\,dy$$ Let’s try some problems:

 Volume Using Washer Method Problem Volume Using Disk Washer Problem Set up and solve the integral that gives the volume of the solid formed by rotating the region about the $$x$$-axis: $$\displaystyle y=1,\,\,\,y=3-\frac{{{{x}^{2}}}}{2}$$   Solution:  Find where the functions intersect: $$\displaystyle 1=3-\frac{{{{x}^{2}}}}{2};\,\,\,\,\,\frac{{{{x}^{2}}}}{2}=2;\,\,\,\,x=\pm 2$$   Now graph. If you’re not sure how to graph, you can always make t-charts. Notice that we have to subtract the volume of the inside function’s rotation from the volume of the outside function’s rotation (move the constant $$\pi$$ to the outside): $$\text{Volume}=\pi \int\limits_{a}^{b}{{\left( {{{{\left[ {R\left( x \right)} \right]}}^{2}}-{{{\left[ {r\left( x \right)} \right]}}^{2}}} \right)}}\,dx$$   Now  integrate to solve: \displaystyle \begin{align}\pi &\int\limits_{{-2}}^{2}{{\left( {{{{\left[ {3-\frac{{{{x}^{2}}}}{2}} \right]}}^{2}}-{{{\left( 1 \right)}}^{2}}} \right)}}\,dx=\pi \int\limits_{{-2}}^{2}{{\left( {9-3{{x}^{2}}+\frac{{{{x}^{4}}}}{4}-1} \right)}}\,dx\\&=\pi \int\limits_{{-2}}^{2}{{\left( {8-3{{x}^{2}}+\frac{{{{x}^{4}}}}{4}} \right)}}\,dx=\pi \left[ {8x-{{x}^{3}}+\frac{{{{x}^{5}}}}{{20}}} \right]_{{-2}}^{2}\,\\&=\pi \left[ {\left( {8\left( 2 \right)-{{2}^{3}}+\frac{{{{2}^{5}}}}{{20}}} \right)-\left( {8\left( {-2} \right)-{{{\left( {-2} \right)}}^{3}}+\frac{{{{{\left( {-2} \right)}}^{5}}}}{{20}}} \right)} \right]\\&=19.2\pi \end{align} Set up (don’t solve) the integral that gives the volume of the solid formed by rotating the region bounded by the equations $$y=x,\,\,\,y=4,\,\,\,x=1$$ about the given lines:  (a) Rotated around the line $$y=5$$ (b) Rotated around the $$y$$-axis   Solutions: Draw first: (a) (b) (a) Since we are rotating around the line $$y=5$$, to get a radius for the “outside” function, which is $$y=x$$, we need to use $$5-x$$ instead of just $$x$$ (try with real numbers and you’ll see). The “inside” part of the washer is the line $$y=5-4=1$$. Thus, the volume is: \begin{align}\pi \int\limits_{a}^{b}{{\left( {{{{\left[ {R\left( x \right)} \right]}}^{2}}-{{{\left[ {r\left( x \right)} \right]}}^{2}}} \right)}}\,dx&=\pi \int\limits_{1}^{4}{{\left( {{{{\left[ {5-x} \right]}}^{2}}-{{1}^{2}}} \right)}}\,dx\\&=\pi \int\limits_{1}^{4}{{\left( {24-10x+{{x}^{2}}} \right)}}\,dx\end{align} (b) Get $$y$$’s in terms of $$x$$. “Outside” function is $$y=x$$, and “inside” function is $$x=1$$. Thus, the volume is: $$\pi \int\limits_{a}^{b}{{\left( {{{{\left[ {R\left( y \right)} \right]}}^{2}}-{{{\left[ {r\left( y \right)} \right]}}^{2}}} \right)}}\,dy=\pi \int\limits_{1}^{4}{{\left( {{{y}^{2}}-{{1}^{2}}} \right)}}\,dy=\pi \int\limits_{1}^{4}{{\left( {{{y}^{2}}-1} \right)}}\,dy$$ Note: It’s coincidental that we integrate up the $$y$$-axis from 1 to 4, like we did across the $$x$$-axis. This is because we are using the line $$y=x$$, so for both integrals, we are going from 1 to 4. Normally the $$y$$ limits would be different than the $$x$$ limits.

# Volumes of Solids: The Shell Method

The shell method for finding volume of a solid of revolution uses integration along an axis perpendicular to the axis of revolution instead of parallel, as we’ve seen with the disk and washer methods. The nice thing about the shell method is that you can integrate around the $$y$$-axis and not have to take the inverse of functions. Also, the rotational solid can have a hole in it (or not), so it’s a little more robust. It’s not intuitive though, since it deals with an infinite number of “surface areas” of rectangles in the shapes of cylinders (shells).

Here are the equations for the shell method:

 Volumes of Solids: The Shell Method Revolution around the $$\boldsymbol {y}$$-axis:       $$\text{Volume}=2\pi \int\limits_{a}^{b}{{x\,f\left( x \right)}}\,dx$$ Revolution around the $$\boldsymbol {x}$$-axis: $$\displaystyle \text{Volume}=2\pi \int\limits_{a}^{b}{{y\,f\left( y \right)}}\,dy$$

Since I believe the shell method is no longer required the Calculus AP tests (at least for the AB test), I will not be providing examples and pictures of this method. Please let me know if you want it discussed further.

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On to Integration by Parts — you are ready!