Integrations

Josef Carl Berthold Puttner (1821-1881) was a German-Austrian landscape and marine painter.

We have sections

\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}

The Definite Integral of a Continuous Function.

Let $f$ be continuous on $[a,b]$. A function $G$ is called an antiderivative for $f$ on $[a,b]$ when $G$ is continuous on $[a,b]$ and $G'(x)=f(x)$ for all $x\in (a,b)$.

Theorem. (The Fundamental Theorem of Integral Calculus). Let $f$ be continuous on $[a,b]$. Suppose that $G$ is any antiderivative for $f$ on $[a,b]$. Then, we have

\[\int_{a}^{b} f(t)dt=G(b)-G(a).\]

Example. Evaluate

\[\int_{0}^{\pi /2} \sin xdx.\]

The antiderivative $G(x)=-\cos x$.

\[\int_{0}^{\pi /2} \sin xdx=-\cos (\pi/2)-[-\cos (1)]=1.\]

Theorem. We have the following properties.

(i) If $f$ is continuous on $[a,b]$ and $a<c<b$, then

\[\int_{a}^{b} f(x)dx=\int_{a}^{c} f(x)dx+\int_{c}^{b} f(x)dx.\]

(ii) The integral of a linear combination is the linear combination of the integrals. In other words, we have

\[\int_{a}^{b} [\alpha f(x)+\beta g(x)]dx=\alpha\int_{a}^{b} f(x)dx+\beta\int_{a}^{b} g(x)dx.\]

Example. We have
\begin{align*}
\int_{0}^{1} (2x-6x^{4}+5)dx & =2\int_{0}^{1} xdx-6\int_{0}^{1} x^{4}dx+\int_{0}^{1} 5dx\\
& =1-\frac{6}{5}+5=4\frac{4}{5}.
\end{align*}

Example. Evaluate

\[\int_{0}^{\pi /4} \sec x(2\tan x-5\sec x)dx.\]

We have

\begin{align*}
\int_{0}^{\pi /4} \sec x(2\tan x-5\sec x)dx & =\int_{0}^{\pi /4} (2\sec x\tan x-5\sec^{2} x)dx\\
& =2\int_{0}^{\pi /4} \sec x\tan xdx-5\int_{0}^{\pi/4} \sec^{2} xdx\\
& =2\left [\sec x\right ]_{0}^{\pi /4}-5\left [\tan x\right ]_{0}^{\pi /4}\\
& =2\sqrt{2}-7.
\end{align*}

If $f$ is nonnegative and continuous on $[a,b]$, then the integral of $f$ from $a$ to $b$ gives the area of the region below the graph of $f$.

\[\mbox{Area = }\int_{a}^{b} f(x)dx.\]

Let $f$ and $g$ be two nonnegative continuous function on $[a,b]$. Suppose that $g(x)\leq f(x)$ for all $x\in [a,b]$. Then the area between $f$ and $g$ is
\[\int_{a}^{b} f(x)dx-\int_{a}^{b} g(x)dx=\int_{a}^{b} [f(x)-g(x)]dx.\]

Example. Find the area of the region bounded above by $y=x+2$ and below by $y=x^{2}$. The first step is to find the points of intersection of the two curves by solving $x+2=x^{2}$, i.e., $(x+1)(x-2)=0$. Therefore, we obtain $x=-1,2$. The curves intersect at $(-1,1)$ and $(2,4)$. The area of the region is given by

\[\int_{-1}^{2} [(x+2)-x^{2}]dx=\frac{9}{2}.\]

As a matter of fact, the assumption that $f$ and $g$ are nonnegative is unnecessary. Suppose that the region $\Omega$ has an upper boundary of the function $y=f(x)$ for all $x\in [a,b]$ and a lower boundary of the function $y=g(x)$ for all $x\in [a,b]$. Then

\[\mbox{Area of $\Omega$ = }\int_{a}^{b} [f(x)-g(x)]dx\]

since we can find a number $c$ such that $f(x)+c\geq 0$ and $g(x)+c\geq 0$. Therefore, we have

\[\int_{a}^{b} [(f(x)+c)-(g(x)+c)]dx=\int_{a}^{b} [f(x)-g(x)]dx.\]

Example. Find the area between the curves $y=\sin x$ and $y=\cos x$. For $\sin x=\cos x$, we obtain $x=\pi /4$ or $5\pi /4$. The area is given by

\[\int_{\pi /4}^{5\pi /4} (\sin x-\cos x)dx=\left [-\cos x-\sin x\right ]_{\pi /4}^{5\pi /4}=2\sqrt{2}.\]

Example. Find the area between the curves $y=4x$ and $y=x^{3}$ from $x=-2$ to $x=2$. We see that $y=x^{3}$ is the upper boundary from $-2$ to $0$, but it is the lower boundary from $0$ to $2$. Therefore, we obtain

\begin{align*}
\mbox{Area } & =\int_{-2}^{0} (x^{3}-4x)dx+\int_{0}^{2} (4x-x^{3})dx\\
& =\left [\frac{1}{4}x^{4}-2x^{2}\right ]_{-2}^{0}+\left [2x^{2}-\frac{1}{4}x^{4}\right ]_{0}^{2}\\
& =8.
\end{align*}

Let $f$ be continuous on $[a,b]$. Then, the integral $\int_{a}^{b} f(x)dx$ can be interpreted as the difference of two areas; that is, the total area of the regions that lie above the $x$-axis minus the total areas of the regions that lie below the $x$-axis. Therefore, the area of region bounded by the graph of $f$ and the $x$-axis for $x\in [a,b]$ is

\[\int_{a}^{b} |f(x)|dx.\]

\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}

Indefinite Integral.

Let $F$ be an antiderivative for $f$. Then we simply write

\[\int f(x)dx=F(x)+C.\]

Antiderivative expressed in this manner are called indefinite integrals. The constant $C$ is called the constant of integration that is an arbitrary constant since it can be assigned any real value. For example, we have

\begin{align*}
& \int x^{r}dx=\frac{x^{r+1}}{r+1}+C\mbox{ ($r$ rational $\neq 1$)}\\
& \int \sin xdx=-\cos x+C\\
& \int \cos xdx=\sin x+C\\
& \int \sec^{2} xdx=\tan x+C\\
& \int \sec x\tan xdx=\sec x+C\\
& \int \csc^{2} xdx=-\cot x+C\\
& \int \csc x\cot xdx=-\csc x+C.
\end{align*}

Let $\alpha$ and $\beta$ be constants. Then, we have

\[\int [\alpha f(x)+\beta g(x)]dx=\alpha\int f(x)dx+\beta\int g(x)dx.\]

Example. Calculate

\[\int [5x^{\frac{3}{2}}-2\csc^{2} x]dx.\]

We have
\begin{align*}
\int [5x^{\frac{3}{2}}-2\csc^{2} x]dx & =5\int x^{\frac{3}{2}}-2\int \csc^{2} xdx\\
& =5(\frac{2}{5})x^{\frac{5}{2}}+C_{1}-2(-\cot x)-C_{2}\\
& =2x^{\frac{5}{2}}+2\cot x+C.
\end{align*}

Example. Find $f$ given that $f”(x)=6x-2$ with $f'(1)=-5$ and $f(1)=3$. We first obtain

\begin{align*} f'(x)+C_{1} & =\int f”(x)dx\\ & =\int (6x-2)dx=3x^{2}-2x+C_{2};\end{align*}

that is, $f'(x)=3x^{2}-2x+C$. Since $f'(1)=-5$, we get $C=-6$. Therefore $f'(x)=3x^{2}-2x-6$. Now, we also get

\begin{align*} f(x)+K_{1} & =\int f'(x)dx\\ & =\int (3x^{2}-2x-6)dx=x^{3}-x^{2}-6x+K_{2};\end{align*}

that is, $f(x)=x^{3}-x^{2}-6x+K$. Since $f(1)=3$, we get $K=9$. Therefore $f(x)=x^{3}-x^{2}-6x+9$.

Change of Variables

An integral of the form

\[\int f(g(x))g'(x)dx\]

can be written as

\[\int f(u)du\]

by setting $u=g(x)$ and $du=g'(x)dx$. If $F$ is an antiderivative for $f$, then, by the chain rule, we have

\[[F(g(x))]’=F'(g(x))g'(x)=f(g(x))g'(x).\]

Therefore, we obtain

\[\int f(g(x))g'(x)dx=\int [F(g(x))]’dx=F(g(x))+C.\]

We can obtain the same result by calculating

\[\int f(u)du\]

and then substituting $g(x)$ back in for $u$ to obtain

\[\int f(u)du=F(u)+C=F(g(x))+C.\]

Example. Calculate

\[\int (x^{2}-1)^{4}2xdx.\]

Set $u=x^{2}-1$. Then $du=2xdx$. Therefore, we have

\begin{align*} \int (x^{2}-1)^{4}2xdx & =\int u^{4}du=\frac{1}{5}u^{5}+C\\ & =\frac{1}{5}(x^{2}-1)^{5}+C.\end{align*}

Example. Calculate

\[\int \sin x\cos xdx.\]

Method 1: Set $u=\sin x$. Then $du=\cos xdx$. Therefore, we have

\begin{align*} \int \sin x\cos xdx & =\int udu=\frac{1}{2}u^{2}+C\\ & =\frac{1}{2}\sin^{2} x+C.\end{align*}

Method 2: Since $\sin x\cos x=\frac{1}{2}\sin 2x$, we have

\[\int \sin x\cos xdx=\frac{1}{2}\int \sin 2xdx.\]

Set $u=2x$ so that $du=2dx$ or $dx=\frac{1}{2}du$. Then, we obtain

\begin{align*} \frac{1}{2}\int \sin 2xdx & =\frac{1}{4}\int \sin udu\\ & =-\frac{1}{4}\cos u+K=-\frac{1}{4}\cos 2x+K.\end{align*}

Note that $\cos 2x=1-2\sin^{2} x$. We also obtain

\begin{align*} -\frac{1}{4}\cos 2x+K & =\frac{1}{2}\sin^{2} x+K-\frac{1}{4}\\ & =\frac{1}{2}\sin^{2} x+C.\end{align*}

Example. Calculate

\[\int x^{2}\sqrt{4+x^{3}}dx.\]

Set $u=4+x^{3}$. Then $du=3x^{2}dx$. Therefore, we have

\begin{align*} \int x^{2}\sqrt{4+x^{3}}dx & =\frac{1}{3}\int \sqrt{u}du=\frac{2}{9}u^{\frac{3}{2}}+C\\ & =\frac{2}{9}(4+x^{3})^{\frac{3}{2}}+C.\end{align*}

Example. Find

\[\int \sec^{3} x\tan xdx.\]

Set $u=\sec x$. Then $du=\sec x\tan xdx$. Therefore, we have

\begin{align*} \int \sec^{3} x\tan xdx & =\int u^{2}du\\ & =\frac{1}{3}u^{3}+C=\frac{1}{3}\sec^{3} x+C.\end{align*}

Example. Find

\[\int x\cos (\pi x^{2})dx.\]

Set $u=\pi x^{2}$. Then $du=2\pi xdx$. Therefore, we have

\begin{align*} \int x\cos (\pi x^{2})dx & =\frac{1}{2\pi}\int \cos udu\\ & =\frac{1}{2\pi}\sin u+C=\frac{1}{2\pi}\sin (\pi x^{2})+C.\end{align*}

Example. Evaluate the definite integral

\[\int_{0}^{2} (x^{2}-1)(x^{3}-3x+2)^{3}dx.\]

Set $u=x^{3}-3x+2$. Then $du=3(x^{2}-1)dx$. Therefore, we obtain

\begin{align*} \int (x^{2}-1)(x^{3}-3x+2)^{3}dx & =\frac{1}{3}\int u^{3}du=\frac{1}{12}u^{4}+C\\ & =\frac{1}{12}(x^{3}-3x+2)^{4}+C.\end{align*}

To evaluate the given definite integral, we need only one antiderivative, and so we will choose the one with $C=0$. This gives
\begin{align*} \int_{0}^{2} (x^{2}-1)(x^{3}-3x+2)^{3}dx & =\left [\frac{1}{12}(x^{3}-3x+2)^{4}\right ]_{0}^{2}\\ & =20.\end{align*}

Example. Find

\[\int x^{2}\csc^{2} x^{3}\cot^{4} x^{3}dx.\]

Method 1: Set $u=x^{3}$. Then $du=3x^{2}dx$. Therefore, we have

\[\int x^{2}\csc^{2} x^{3}\cot^{4} x^{3}dx=\frac{1}{3}\int\cot^{4} u\csc^{2} udu.\]

Set $t=\cot u$. Then $dt=-\csc^{2} udu$. Therefore, we have

\begin{align*} \frac{1}{3}\int \cot^{4} u\csc^{2} udu & =-\frac{1}{3}\int t^{4}dt\\ & =-\frac{1}{15}t^{5}+C=-\frac{1}{15}\cot^{5} u+C,\end{align*}

which implies

\[\int x^{2}\csc^{2} x^{3}\cot^{4} x^{3}dx=-\frac{1}{15}\cot^{5} x^{3}+C.\]

Method 2: Set $u=\cot x^{3}$. Then $du=-\csc^{2} x^{3}\cdot 3x^{2}dx$. Therefore, we have

\begin{align*} \int x^{2}\csc^{2} x^{3}\cot^{4} x^{3}dx & =-\frac{1}{3}\int u^{4}du\\ & =-\frac{1}{15}u^{5}+C=-\frac{1}{15}\cot^{5} x^{3}+C.\end{align*}

Theorem. (Change of Variables for Definite Integral). Suppose that $g'(x)$ is continuous on $[a,b]$, and $f$ is continuous on the set of values taken on by $g$. Then, we have

\[\int_{a}^{b} f(g(x))g'(x)dx=\int_{g(a)}^{g(b)} f(u)du.\]

Proof. Let $F$ be antiderivative for $f$. Then, we have

\begin{align*}
\int_{a}^{b} f(g(x))g'(x)dx & =\int_{a}^{b} F'(g(x))g'(x)dx\\
& =F(g(b))-F(g(a))\\
& =\int_{g(a)}^{g(b)} f(u)du.
\end{align*}
This completes the proof. $\blacksquare$

We also note that

\[\int_{a}^{b} f(x)dx=-\int_{b}^{a} f(x)dx.\]

Example. Evaluate

\[\int_{0}^{1/2} \cos^{3} \pi x\sin \pi xdx.\]

Set $u=\cos \pi x$. Then $du=-\pi\sin \pi xdx$. At $x=0$, we have $u=1$. At $x=1/2$, we have $u=0$. Therefore, we obtain

\begin{align*} \int_{0}^{1/2} \cos^{3} \pi x\sin \pi xdx & =-\frac{1}{\pi}\int_{1}^{0}u^{3}du\\ & =\frac{1}{\pi}\int_{0}^{1} u^{3}du\\ & =\frac{1}{\pi}\left [\frac{1}{4}u^{4}\right ]_{0}^{1}\\ & =\frac{1}{4\pi}.\end{align*}

Example. Evaluate

\[\int_{0}^{\sqrt{3}} x^{5}\sqrt{x^{2}+1}dx.\]

Set $u=x^{2}+1$. Then $du=2xdx$. At $x=0$, we have $u=1$. At $x=\sqrt{3}$, we have $u=4$. Therefore, we obtain

\begin{align*}
\int_{0}^{\sqrt{3}} x^{5}\sqrt{x^{2}+1}dx & =\frac{1}{2}\int_{1}^{4}(u-1)^{2}\sqrt{u}du\\
& =\frac{1}{2}\int_{1}^{4} (u^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}})du\\
& =\frac{1}{2}\left [\frac{2}{7}u^{\frac{7}{2}}-\frac{4}{5}u^{\frac{5}{2}}+\frac{2}{3}u^{\frac{3}{2}}\right ]_{1}^{4}\\
& =\frac{848}{105}.
\end{align*}

Theorem. We have the following properties.

(i) If $f(x)\geq 0$ for all $x\in [a,b]$, then

\[\int_{a}^{b} f(x)dx\geq 0.\]

(ii) If $f(x)>0$ for all $x\in [a,b]$, then

\[\int_{a}^{b} f(x)dx>0.\]

(iii) If $f(x)\leq g(x)$ for all $x\in [a,b]$, then

\[\int_{a}^{b} f(x)dx\leq\int_{a}^{b} g(x)dx.\]

(iv) If $f(x)<g(x)$ for all $x\in [a,b]$, then

\[\int_{a}^{b} f(x)dx<\int_{a}^{b} g(x)dx.\]

(v) We have

\[\left |\int_{a}^{b} f(x)dx\right |\leq\int_{a}^{b} |f(x)|dx.\]

(vi) If $m$ is the minimum value of $f$ on $[a,b]$ and $M$ is the maximum value, then

\[m(b-a)\leq\int_{a}^{b} f(x)dx\leq M(b-a).\]

(vii) If $u$ is a differentiable function of $x$ and $f$ is continuous, then

\[\frac{d}{dx}\left (\int_{a}^{u} f(t)dt\right )=f(u)\frac{du}{dx}.\]

for any constant $a$.

(vii) If $f$ is odd on $[-a,a]$, then

\[\int_{-a}^{a} f(x)dx=0.\]

(viii) If $f$ is even on $[-a,a]$, then

\[\int_{-a}^{a} f(x)dx=2\int_{0}^{a} f(x)dx.\]

Example. Find

\[\frac{d}{dx}\left (\int_{0}^{x^{3}} \frac{dt}{1+t}\right ).\]

We have

\begin{align*} \frac{d}{dx}\left (\int_{0}^{x^{3}} \frac{dt}{1+t}\right ) & =f(u)\frac{du}{dx}\\ & =\frac{1}{1+x^{3}}\cdot 3x^{2}=\frac{3x^{2}}{1+x^{3}}.\end{align*}

Example. Find

\[\frac{d}{dx}\left (\int_{x}^{2x} \frac{dt}{1+t^{2}}\right ).\]

Since

\[\int_{0}^{x} \frac{dt}{1+t^{2}}+\int_{x}^{2x} \frac{dt}{1+t^{2}}=\int_{0}^{2x} \frac{dt}{1+t^{2}},\]

we have

\[\int_{x}^{2x} \frac{dt}{1+t^{2}}=\int_{0}^{2x} \frac{dt}{1+t^{2}}-\int_{0}^{x} \frac{dt}{1+t^{2}}.\]

Differentiation gives

\begin{align*}
\frac{d}{dx}\left (\int_{x}^{2x} \frac{dt}{1+t^{2}}\right )
& =\frac{d}{dx}\left (\int_{0}^{2x} \frac{dt}{1+t^{2}}\right )-\frac{d}{dx}\left (\int_{0}^{x} \frac{dt}{1+t^{2}}\right )\\
& =\frac{2}{1+4x^{2}}-\frac{1}{1+x^{2}}.
\end{align*}

Theorem. (The First Mean-Value Theorem for Integrals). Suppose that $f$ is continuous on $[a,b]$. Then, there is at least one number $c$ in $(a,b)$ satisfying

\[\int_{a}^{b} f(x)dx=f(c)(b-a).\]

This number $f(c)$ is called the average (or mean) value of $f$ on $[a,b]$.

Theorem. (The Second Mean-Value Theorem for Integrals). Suppose that $f$ and $g$ are continuous on $[a,b]$ and $g$ is nonnegative. Then, there
is a number $c$ in $(a,b)$ satisfying

\[\int_{a}^{b} f(x)g(x)dx=f(c)\int_{a}^{b} g(x)dx.\]

This number $f(c)$ is called the $g$-weighted average of $f$ on $[a,b]$.

\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}

Some Applications of the Integral.

Example. Find the area of the region bounded on the left by $x=y^{2}$ and on the right by $x=3-2y^{2}$. First of all, we have to solve the points of intersection. Now $y^{2}=3-2y^{2}$ implies $y=\pm 1$. Therefore, the points of intersection are $(1,1)$ and $(1,-1)$. The area is given by

\begin{align*} \int_{-1}^{1} [(3-2y^{2})-y^{2}]dy & =\int_{-1}^{1} (3-3y^{2})dy\\ & =\left [3y-y^{3}\right ]_{-1}^{1}=4.\end{align*}

Example. Calculate the area of the region bounded by the curves $x=y^{2}$ and $x-y=2$ by integrating (a) with respect to $x$, (b) with respect to $y$. We can verify that the two curves intersect at the points $(1,-1)$ and $(4,2)$.

(a) The upper boundary of the region is the curve $y=\sqrt{x}$. However, the lower boundary is described by two different equations: $y=-\sqrt{x}$ from $x=0$ to $x=1$, and $y=x-2$ from $x=1$ to $x=4$. Thus, we need two integrals

\begin{align*}
\int_{0}^{1} [\sqrt{x}-(-\sqrt{x})]dx+\int_{1}^{4} [\sqrt{x}-(x-2)]dx & =2\int_{0}^{1} \sqrt{x}dx+\int_{1}^{4} (\sqrt{x}-x+2)dx\\
& =\left [\frac{4}{3}x^{\frac{3}{2}}\right ]_{0}^{1}+\left [\frac{2}{3}x^{\frac{3}{2}}-\frac{1}{2}x^{2}+2x\right ]_{1}^{4}=\frac{9}{2}.
\end{align*}

(b) Now the right boundary is the line $x=y+2$ and the left boundary is the curve $x=y^{2}$. Since $y$ ranges from $-1$ to $2$, the area is given by

\begin{align*} \int_{-1}^{2} [(y+2)-y^{2}]dy & =\left [\frac{1}{2}y^{2}+2y-\frac{1}{3}y^{3}\right ]_{-1}^{2}\\ & =\frac{9}{2}.\end{align*}

We suppose that the solid lies entirely between $x=a$ and $x=b$. We denote by $A(x)$ the area of the cross section that has coordinate $x$. If the cross-sectional area, $A(x)$, varies continuously with $x$, then we can find the volume of the solid by integrating $A(x)$ from $a$ to $b$:

\[V=\int_{a}^{b} A(x)dx.\]

Example. Find the volume of a pyramid whose altitude is $h$ and whose base is a square with sides of length $r$. We can get

\[A(x)=\frac{r^{2}(h-x)^{2}}{h^{2}}.\]

and

\begin{align*}
V=\int_{0}^{h} A(x)dx & =\frac{r^{2}}{h^{2}}\int_{0}^{h} (h-x)^{2}dx\\
& =\frac{r^{2}}{h^{2}}\left [-\frac{(h-x)^{3}}{3}\right ]_{0}^{h}=\frac{1}{3}r^{2}h.
\end{align*}

Example. The base of a solid is the region bounded by the ellipse

\[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1.\]

Find the volume of the solid given that each cross section perpendicular to the $x$-axis is an isosceles triangle with base in the region and altitude equal to one-half the base. We can get

\begin{align*} A(x) & =\frac{1}{2}bh\\ & =\frac{1}{2}\left (\frac{2b}{a}\sqrt{a^{2}-x^{2}}\right )\left (\frac{b}{a}\sqrt{a^{2}-x^{2}}\right )\\ & =\frac{b^{2}}{a^{2}}(a^{2}-b^{2}).\end{align*}

The volume is

\begin{align*}
V & =\int_{-a}^{a} A(x)dx\\
& =2\int_{0}^{a} A(x)dx\mbox{ (by symmetry)}\\
& =\frac{2b^{2}}{a^{2}}\int_{0}^{a} (a^{2}-x^{2})dx\\
& =\frac{2b^{2}}{a^{2}}\left [a^{2}x-\frac{x^{3}}{3}\right ]_{0}^{a}\\
& =\frac{4}{3}ab^{2}.
\end{align*}

Disc Method.

Suppose that $f$ is nonnegative and continuous on $[a,b]$. If we revolve the region bounded by the graph of $f$ and $x$-axis about the $x$-axis, we obtain a solid. The volume of this solid is given by the formula

\[V=\int_{a}^{b} \pi [f(x)]^{2}dx.\]

Example. We can generate a cone of base radius $r$ and height $h$ by revolving about the $x$-axis the region below the graph of

\[f(x)=\frac{rx}{h}\mbox{ for }0\leq x\leq h.\]

Therefore, the volume is

\begin{align*} V & =\int_{0}^{h} \pi \left [\frac{rx}{h}\right ]^{2}dx\\ & =\frac{\pi r^{2}}{h^{2}}\int_{0}^{h} x^{2}dx=\frac{1}{3}\pi r^{2}h.\end{align*}

Example. A sphere of radius $r$ can be obtained by revolving about the $x$-axis the region below the graph of

\[f(x)=\sqrt{r^{2}-x^{2}}\mbox{ for }-r\leq x\leq r.\]

Therefore, the volume is

\begin{align*} V & =\int_{r}^{-r} \pi (r^{2}-x^{2})dx\\ & =\pi\left [r^{2}x-\frac{1}{3}x^{3}\right ]_{-r}^{r}=\frac{4}{3}\pi r^{3}.\end{align*}

Now suppose that $x=g(y)$ is continuous and nonnegative for $c\leq y\leq d$. If we revolve the region bounded by the graph of $g$ and the $y$-axis about
the $y$-axis. Then the volume of this solid is given by

\[V=\int_{c}^{d} \pi [g(y)]^{2}dy.\]

Example. Find the volume of the solid generated by revolving the region bounded by $y=x^{2/3}+1$, $0\leq x\leq 8$, about the $y$-axis. If $y=x^{2/3}+1$ for $0\leq x\leq 8$, then $x=(y-1)^{3/2}$ for $1\leq y\leq 5$. Therefore, the volume is

\begin{align*} V & =\int_{1}^{5} \pi [g(y)]^{2}dy=\pi\int_{1}^{5} (y-1)^{3}dy\\ & =\pi\left [\frac{(y-1)^{4}}{4}\right ]_{1}^{5}=64\pi .\end{align*}

Suppose that $f$ and $g$ are nonnegative continuous functions with $g(x)\leq f(x)$ for all $x\in [a,b]$. If we revolve the region about the $x$-axis, we obatin a solid. The volume of this solid is given by the formula

\[V=\int_{a}^{b} \pi ([f(x)]^{2}-[g(x)]^{2})dx.\]

Suppose now that the boundaries are functions of $y$ rather than $x$. By revolving the region about $y$-axis, we obtain a solid. The volume of this solid is given by the formula

\[V=\int_{c}^{d} \pi ([F(y)]^{2}-[G(y)]^{2})dy.\]

Example. Find the volume of the solid generated by revolving the region between $y=x^{2}$ and $y=2x$ about the $x$-axis. Setting $x^{2}=2x$, we get $x=0,2$. Thus the curves intersect at the points $(0,0)$ and $(2,4)$. Then, we have

\begin{align*} V & =\int_{0}^{2} \pi [(2x)^{2}-(x^{2})^{2}]dx\\ & =\pi\int_{0}^{2}(4x^{2}-x^{4})dx=\frac{64}{15}\pi.\end{align*}

Example. Find the volume of the solid generated by revolving the region between $y=x^{2}$ and $y=2x$ about the $y$-axis. We have

\begin{align*} V & =\int_{0}^{4} \pi [(\sqrt{y})^{2}-(\frac{1}{2}y)^{2}]dy\\ & =
\pi\int_{0}^{4} (y-\frac{1}{4}y^{2})dy=\frac{8}{3}\pi.\end{align*}

Shell Method.

Let $[a,b]$ be an interval that does not contain $0$ in its interior, and let $f$ be a nonnegative continuous function on $[a,b]$. If the region bounded by the graph of $f$ and the $x$-axis is revolved around the $y$-axis, then a solid is generated. The volume of this solid is given by the shell method formula

\[V=\int_{a}^{b} 2\pi xf(x)dx.\]

Example. The region bounded by $f(x)=4x-x^{2}$ and the $x$-axis between $x=1$ and $x=4$ is rotated about $y$-axis. Find the volume of the solid that is
generated. Then

\[V=\int_{1}^{4} 2\pi x(4x-x^{2})dx=\frac{81}{2}\pi.\]

Suppose that $f$ and $g$ are nonnegative continuous functions with $g(x)\leq f(x)$ for all $x\in [a,b]$. If we revolve the region about the $y$-axis, we obtain a solid. The volume of this solid is given by the formula

\[V=\int_{a}^{b} 2\pi x[f(x)-g(x)]dx.\]

We can also apply the shell method to solids generated by revolving a region about the $x$-axis. In this instance the curved boundaries are functions of $y$ rather than $x$. Then, we have

\[V=\int_{c}^{d} 2\pi y[F(y)-G(y)]dy.\]

Example. Find the volume of the solid generated by revolving the region between $y=x^{2}$ and $y=2x$ about the $y$-axis. We have

\[V=\int_{0}^{2} 2\pi x(2x-x^{2})dx=\frac{8}{3}\pi.\]

Example. Find the volume of the solid generated by revolving the region between $y=x^{2}$ and $y=2x$ about the $x$-axis. We have

\[V=\int_{0}^{4} 2\pi y(\sqrt{y}-\frac{1}{2}y)dy=\frac{65}{14}\pi.\]

Example. A round hole of radius $r$ is drilled through the center of a hemisphere of radius $a$ ($r<a$). Find the volume of the portion of the hemisphere that remains. A hemisphere of radius $a$ can be generated by revolving the region in the first quadrant bounded by $x^{2}+y^{2}=a^{2}$ around the $y$-axis.

Method 1: we have

\begin{align*} V & =\int_{0}^{\sqrt{a^{2}-r^{2}}}\left (\pi\left [\sqrt{a^{2}-x^{2}}
\right ]^{2}-\pi r^{2}\right )dy\\ & =\pi\int_{0}^{\sqrt{a^{2}-r^{2}}}
(a^{2}-r^{2}-y^{2})dy\\ & =\frac{2}{3}\pi(a^{2}-r^{2})^{\frac{3}{2}}.\end{align*}

Method 2: we have

\[V=\int_{r}^{a} 2\pi x\sqrt{a^{2}-x^{2}}dx.\]

Let $u=a^{2}-x^{2}$. Then $du=-2xdx$. We have $u=a^{2}-r^{2}$ at $x=r$, and $u=0$ at $x=a$. Therefore, we obtain

\begin{align*} V & =\int_{r}^{a} 2\pi x\sqrt{a^{2}-x^{2}}dx\\ & =-\pi\int_{a^{2}-r^{2}}^{0}
u^{\frac{1}{2}}du\\ & =\pi\int_{0}^{a^{2}-r^{2}} u^{\frac{1}{2}}du\\ & =
\frac{2}{3}\pi(a^{2}-r^{2})^{\frac{3}{2}}.\end{align*}

Example. The region between $y=\sqrt{x}$ and $y=x^{2}$ for $0\leq x\leq 1$ is revolved around the line $x=-2$. Find the volume of the solid which is generated. Using the shell method, for each $x\in [0,1]$, the line segment at $x$ generates a cylinder of radius $x+2$, height $\sqrt{x}-x^{2}$. Therefore, the volume is given by

\begin{align*} V & =\int_{0}^{1} 2\pi (x+2)(\sqrt{x}-x^{2})dx\\ & =2\pi\int_{0}^{1}
(x^{\frac{3}{2}}+2x^{\frac{1}{2}}-x^{3}-2x^{2})dx\\ & =\frac{49}{30}\pi.\end{align*}