# 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 $y=x^2, \quad y=x$ 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{aligned} y&=x^2\\ y&=x. \end{aligned} 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 $$\mathrm{d} y$$. And the volume of the washer is approximately $\pi(y-y^2)\mathrm{d} y.$

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{aligned} V&=\int_0^1 \pi(y-y^2)\mathrm{d} y\\ &= \pi\int_0^1(y-y^2)\mathrm{d} y\\ &=\pi(\frac{y^2}{2}-\frac{y^3}{3})|_0^1\\ &=\pi(\frac12-\frac13)\\ &=\frac{\pi}{6}. \end{aligned}

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

Solution.

Solve for $$x$$ from the equation $\frac14 x^2=5-x^2.$ 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 $$|y_I -0|=\frac14x^2$$ and the outer radius is $$|y_O-0|=5-x^2$$. Because $$5-x^2\geq \frac14 x^2$$ over $$[-2, 2]$$.

So the area of the annulus is $\pi y_O^2- \pi y_I^2=\pi\left((5-x^2)-\left(\frac14x^2\right)^2\right).$ The volume of the washer is approximately $\pi((5-x^2)^2-(\frac14x^2)^2)\mathrm{d} x.$

So the volume of the solid is \begin{aligned} V&=\int_{-2}^2 \pi\left((5-x^2)-\left(\frac14x^2\right)^2\right)\mathrm{d} x\\ &=\frac{5\pi}{16} \int_{-2}^2(3x^4-32x^2+80)\mathrm{d} x\\ &=\frac{176\pi}{3}. \end{aligned}

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

Solution.
Since the rotation axis is horizontal and $$y$$-values are easy to obtain, we use washer method. Note that $$\cos x\geq \sin x\geq 0$$ 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{aligned} V&=\int_0^{\pi/4} \pi((\cos x+1)^2-(\sin x +1)^2)\mathrm{d} x\\ &= \pi\int_0^{\pi/4}(\cos(2x)+2\cos x-2\sin x)\mathrm{d} x\\ &=\pi+4\pi\sqrt{2} \end{aligned}

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 $$\cos^2 x\leq 1$$ 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{aligned} V&=\int_{\pi/2}^{\pi/2} \pi(1-(1-\cos^2 x)^2)\mathrm{d} x\\ &= 2\pi \int_0^{\pi/2}(2\cos^2 x-\cos^4 x)\mathrm{d} x\\ &=\frac{5\pi^2}{8} \end{aligned}

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

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

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{aligned} V&=\int_0^{1}(2\sqrt{1-y})^2 \mathrm{d} y\\ &= 2. \end{aligned}

## 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 $y=\sqrt{x}, \quad y=0, \quad x=1$ 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 $$0\le \sqrt{x}\le 1$$ for $$0\le x\le 1$$, the height is $$|y_T-y_B|=|y-0|=\sqrt{x}$$. The thickness is $$\mathrm{d} x$$, 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{aligned} V&=\int_0^{1} 2\pi x \sqrt{x}\mathrm{d} x\\ &= 2\pi \int_0^{1} x^{4/3} \mathrm{d} x\\ &=\frac{6\pi}{7}. \end{aligned}

Exercise 6.7 Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves $y=x^{2}, \quad y=6x-2x^{2}$ 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 $$6x-2x^2\geq x^2$$ over $$[0, 2]$$. The radius of the shell over $$(x, 0)$$ is $$x$$. With the thickness $$\mathrm{d}x$$, the volume of the cylindrical shell over $$(x, 0)$$ is approximately $$2\pi x(6x-2x^2)\mathrm{d}x$$.

The volume of the solid is \begin{aligned} V&=\int_0^2 2\pi x(6x-2x^2)\mathrm{d} x\\ &= 2\pi\int_0^2 (6x^2-2x^3)\mathrm{d} x\\ &=8\pi \end{aligned}

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

Solution.

The thickness is $$\mathrm{d} y$$ 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{aligned} V&=\int_0^2 2\pi y\cdot y^2\mathrm{d} y\\ &= 2\pi\int_0^{2} y^3\mathrm{d} y\\ &=8\pi. \end{aligned}

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

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{aligned} V&=\int_0^3 (2\pi y) ((4-(y-1)^2)-(3-y)) \mathrm{d} y\\ &=2\pi \int_0^3 (3y^2 - y^3) \mathrm{d} y\\ &=\frac{27\pi}{2}. \end{aligned}

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{aligned} V&=\int_0^2 \pi x^2 \mathrm{d} y\\ &=\pi\int_0^2 (1-(y-1)^2) \mathrm{d} y\\ &=\frac{4\pi}{3}. \end{aligned}