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

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