Topic 6 Application of Integration

6.1 Volumes

Exercise 6.1 Find the volume of the solid obtained by rotating the region enclosed by the given curves $$y=x^2,y=x$$ about $y$-axis.

Solution.

We start by find the intersections of the curves which are the conner points of the region. The intersections are the solutions the system of equations $$\begin{array}{cc}y& =x^2\\ y& =x.\end{array}$$ Solve this system, we get $(0,0)$ and $(1,1)$.

Remark: The intersection of curves are the conner points of the region which can be helpful to do calculation without drawing.

Take a cross-section over $(0,y)$ with respect to the rotation axis, the $y$-axis, you will see an annulus.

Remark: A cross-section is perpendicular to the rotation axis.

Because the region is over $0\le y\le 1$ and $y\le \sqrt{y}$ for $0\le y\le 1$. The inner radius is $|x_I-0|=y$ and the outer radius is $|x_O-0|=\sqrt{y}$. So the area of the annulus is $$\pi x_O^2-\pi x_I^2=\pi (y-y^2).$$ The solid can be considered as created by moving the annulus along $y$-axis. Take a small change in $y$ direction, we get a washer with the thickness $dy$. And the volume of the washer is approximately $$\pi (y-y^2)dy.$$

Remark: The cross-section intersects the region along a line segment with endpoints $(x_I,y)$ and $(x_O,y)$.
The inequality $y\le \sqrt{y}$ can be checked using a test point in the interval $[0,1]$. (Do you know why?)

So the volume of the solid is $$\begin{array}{cc}V& ={\int }_0^1\pi (y-y^2)dy\\ & =\pi {\int }_0^1(y-y^2)dy\\ & =\pi (\frac{y^2}{2}-\frac{y^3}{3}){|}_0^1\\ & =\pi (\frac{1}{2}-\frac{1}{3})\\ & =\frac{\pi }{6}.\end{array}$$

Exercise 6.2 Find the volume of the solid obtained by rotating the region bounded by the given curves $$y=\frac{1}{4}{x}^2,y=5-x^2$$ about the $x$-axis

Solution.

Solve for $x$ from the equation $$\frac{1}{4}{x}^2=5-x^2.$$ We get $x=-2$ or $x=2$. The region has two conner points $(-2,1)$ and $(2,1)$.

The cross-section through $(x,0)$ and perpendicular to the rotation axis, the $x$-axis, is an annulus. The inner radius is $|y_I-0|=\frac{1}{4}{x}^2$ and the outer radius is $|y_O-0|=5-x^2$. Because $5-x^2\ge \frac{1}{4}{x}^2$ over $[-2,2]$.

So the area of the annulus is $$\pi y_O^2-\pi y_I^2=\pi \left(\right(5-x^2)-{\left(\frac{1}{4}{x}^2\right)}^2).$$ The volume of the washer is approximately $$\pi \left(\right(5-x^2{)}^2-\left(\frac{1}{4}{x}^2{)}^2\right)dx.$$

So the volume of the solid is $$\begin{array}{cc}V& ={\int }_{-2}^2\pi \left(\right(5-x^2)-{\left(\frac{1}{4}{x}^2\right)}^2)dx\\ & =\frac{5\pi }{16}{\int }_{-2}^2(3x^4-32x^2+80)dx\\ & =\frac{176\pi }{3}.\end{array}$$

Exercise 6.3 Find the volume of the solid obtained by rotating the region bounded by the given curves $$y=\sin x,y=\cos x,0\le x\le \pi /4,$$ about $y=-1$.

Solution.
Since the rotation axis is horizontal and $y$-values are easy to obtain, we use washer method. Note that $\cos x\ge \sin x\ge 0$ for $0\le x\le \pi /4$, then the cross-section through a point $(x,0)$ has the inner radius $r_I=|\sin x-(-1)|=\sin x+1$ and outer radius $r_O=|\cos x-(-1)|=\cos x+1$. Then the volume of the solid is $$\begin{array}{cc}V& ={\int }_0^{\pi /4}\pi \left(\right(\cos x+1{)}^2-(\sin x+1{)}^2)dx\\ & =\pi {\int }_0^{\pi /4}\left(\cos \right(2x)+2\cos x-2\sin x)dx\\ & =\pi +4\pi \sqrt{2}\end{array}$$

Exercise 6.4 Find the volume of the solid obtained by rotating the region bounded by the given curves $$y=0,y={\cos }^{2}x,-\pi /2\le x\le \pi /2,$$ about $y=1$.

Solution.

Note that ${\cos }^{2}x\le 1$ for $-\pi /2\le x\le \pi /2$, then the cross-section through a point $(x,0)$ has the inner radius $r_I=|{\cos }^{2}x-1|=1-{\cos }^{2}x$ and outer radius $r_O=|0-1|=1$. Then the volume of the solid is $$\begin{array}{cc}V& ={\int }_{\pi /2}^{\pi /2}\pi (1-(1-{\cos }^{2}x{)}^2)dx\\ & =2\pi {\int }_0^{\pi /2}(2{\cos }^{2}x-{\cos }^{4}x)dx\\ & =\frac{5{\pi }^2}{8}\end{array}$$

Remark. One way to evaluate the integral is to use the reduction formula $\int {\cos }^{m}xdx=dfrac\sin x{\cos }^{m-1}xm+dfracm-1m\int {\cos }^{m-2}xdx,$ which can be obtained by integration by parts.

Another way is to use (twice) the double angle formula: $\cos \left(2x\right)=2{\cos }^{2}x-1.$

Exercise 6.5 Find the volume of the described solid $S$. The base of is the region enclosed by the parabola $y=1-x^2$ and the $x$-axis. Cross-sections perpendicular to the $y$-axis are squares.

Solution.

To find the volume, we need find the area of the square, equivalently, the side length of the square. Since cross-sections are perpendicular to the $y$-axis, the cross-section passing through $(0,y)$ intersects the enclosed region along a line segment parallel to the $x$-axis. This line segment has the same length as a side of the square. The length of the line segment is given by $$|x_1-x_2|=|\sqrt{1-y}-(-\sqrt{1-y})|=2\sqrt{1-y}.$$

The region bounded by $y=0$ and $y=1$, the maximum of the function. Then the volume of the solid is $$\begin{array}{cc}V& ={\int }_0^1(2\sqrt{1-y}{)}^{2}dy\\ & =2.\end{array}$$

6.2 Volumes by Cylindrical Shells

Exercise 6.6 Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves $$y=\sqrt[3]{3x},y=0,x=1$$ about the $y$-axis.

Solution.

As the rotation axis is the $y$-axis, the cylindrical shell above $(x,0)$ is parallel to the $y$-axis with radius approximately $x$. The height of a shell is the distance of $y$-coordinate on the boundary of the region above $x$. Since $0\le \sqrt[3]{3x}\le 1$ for $0\le x\le 1$, the height is $|y_T-y_B|=|y-0|=\sqrt[3]{3x}$. The thickness is $dx$, because the cylindrical shells extend along the $x$-axis and away from the rotation axis $y$-axis.

So the volume of the solid is $$\begin{array}{cc}V& ={\int }_0^{1}2\pi x\sqrt[3]{3x}dx\\ & =2\pi {\int }_0^1{x}^{4/3}dx\\ & =\frac{6\pi }{7}.\end{array}$$

Exercise 6.7 Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves $$y=x^2,y=6x-2x^2$$ about the $y$-axis.

Solution.

By solving a system of equations, we find the curves intersect at $(0,0)$ and $(2,4)$.

Since the rotation axis is the $y$-axis. Cylindrical shells are parallel to the $y$-axis. The height of a shell over $(x,0)$ is $|y_T-y_B|=|(6x-2x^2)-x^2|=6x-3x^2$. Because $6x-2x^2\ge x^2$ over $[0,2]$. The radius of the shell over $(x,0)$ is $x$. With the thickness $dx$, the volume of the cylindrical shell over $(x,0)$ is approximately $2\pi x(6x-2x^2)dx$.

The volume of the solid is $$\begin{array}{cc}V& ={\int }_0^{2}2\pi x(6x-2x^2)dx\\ & =2\pi {\int }_0^2(6x^2-2x^3)dx\\ & =8\pi \end{array}$$

Exercise 6.8 Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves $$y=\sqrt{x},x=0,y=2$$ about the $x$-axis.

Solution.

The thickness is $dy$ because the rotation axis is the $x$-axis. The radius is the distance away from the $x$-axis, which is $y$. The height is the distance of the boundaries of the region above $y$ which is $|x_R-x_L|=x=y^2$.

So the volume of the solid is $$\begin{array}{cc}V& ={\int }_0^{2}2\pi y\cdot y^{2}dy\\ & =2\pi {\int }_0^2{y}^{3}dy\\ & =8\pi .\end{array}$$

Exercise 6.9 Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves $$x+y=3,x=4-(y-1{)}^2$$ about the $x$-axis.

Solution.

By solving the system of the two equations, we get the conner points of the region $(3,0)$ and $(0,3)$. Over $(0,y)$ for any $0\le y\le 3$, we know that $4-(y-1{)}^2\ge 3-y$.

Consider the shell over the point $(0,y)$. The radius is $y$. The height is $|x_R-x_L|=(4-(y-1{)}^2)-(3-y)$

The volume of the solid is $$\begin{array}{cc}V& ={\int }_0^3\left(2\pi y\right)\left(\right(4-(y-1{)}^2)-(3-y))dy\\ & =2\pi {\int }_0^3(3y^2-y^3)dy\\ & =\frac{27\pi }{2}.\end{array}$$

Exercise 6.10 Find the volume generated by rotating the region bounded by the given curves $$x^2+(y-1{)}^2=1$$ about the $y$-axis.

Solution.

To use shell method, we will need to solve for $y$ to get the height which will involve radicals.

Since the rotation axis is the $y$-axis, using the disk method, we will need to find $\pi x^2$ which is the area of the cross-section (a disk) over $(0,y)$.

So the disk method seems easier.

Since $(y-1{)}^2=1-x^2\le 1$, solving the inequality for $y$, we get $0\le y\le 2$. Then the volume of the solid is $$\begin{array}{cc}V& ={\int }_0^2\pi x^{2}dy\\ & =\pi {\int }_0^2(1-(y-1{)}^2)dy\\ & =\frac{4\pi }{3}.\end{array}$$