Perhatikan gambar berikut!

$\begin{aligned}V & =\pi\int_{0}^{2}\left(3x\right)^{2}dx\\
& =\pi\int_{0}^{2}9x^{2}dx\\
& =\pi\left[3x^{3}\right]_{0}^{2}\\
& =24\pi.
\end{aligned}
$

Perhatikan gambar berikut!

$\begin{aligned}V & =\pi\int_{0}^{2}\left(x+1\right)^{2}dx\\
& =\pi\int_{0}^{2}\left(x^{2}+2x+1\right)dx\\
& =\pi\left[\frac{1}{3}x^{3}+x^{2}+x\right]_{0}^{2}
\end{aligned}
$

$\begin{aligned}V & =\pi\left(\frac{8}{3}+4+2\right)\\
& =\pi\left(6+\frac{8}{3}\right)\\
& =\frac{26}{3}\pi.
\end{aligned}
$

$\begin{aligned}V & =\pi\int_{1}^{3}\left(4-x\right)^{2}dx\\
& =\pi\int_{1}^{3}\left(16-8x+x^{2}\right)dx\\
& =\pi\left[16x-4x^{2}+\frac{1}{3}x^{3}\right]_{1}^{3}
\end{aligned}
$

$\begin{aligned}V & =\pi\left[\left(48-36+9\right)-\left(16-4+\frac{1}{3}\right)\right]\\
& =\pi\left(21-12-\frac{1}{3}\right)\\
& =\pi\left(9-\frac{1}{3}\right)\\
& =\frac{26}{3}\pi\\
& =8\frac{2}{3}\pi.
\end{aligned}
$

$x-y^{2}+1=0$

$\rightarrow y^{2}=x+1$

$\begin{aligned}V & =\pi\int_{-1}^{4}y^{2}dx\\
& =\pi\int_{-1}^{4}(x+1)dx
\end{aligned}
$

$\begin{aligned}V & =\pi\left[\frac{1}{2}x^{2}+x\right]_{-1}^{4}\\
& =\pi\left\{ \left(8+4\right)-\left(\frac{1}{2}-1\right)\right\} \\
& =\pi\left(12+\frac{1}{2}\right)\\
& =12\frac{1}{2}\pi.
\end{aligned}
$

Cari titik perpotongan antara kurva $y=x^{2}$ dan garis $y=x+2$

$x^{2}=x+2$

$x^{2}-x-2=0$

$\left(x-2\right)(x+1)=0$

$x=2$ atau $x=-1$

Perhatikan gambar berikut!

$V=\pi\int_{-1}^{2}\left\{ \left(x+2\right)^{2}-\left(x^{2}\right)^{2}\right\} dx$

$V=\pi\int_{-1}^{2}\left(-x^{4}+x^{2}+4x+4\right)dx$

$=\pi\left[-\frac{1}{5}x^{5}+\frac{1}{3}x^{3}+2x^{2}+4x\right]_{-1}^{2}$

$V=\pi\left\{ \left(-\frac{32}{5}+\frac{8}{3}+8+8\right)-\left(\frac{1}{5}-\frac{1}{3}+2-4\right)\right\} $

$=\pi\left(-\frac{33}{5}+\frac{9}{3}+16+2\right)$

$=\pi\left(21-\frac{33}{5}\right)$

$=\frac{72}{5}\pi$

$=14\frac{2}{5}\pi.$

Cari titik potong kurva $y=1-x^{2}$ dan garis $y=1-x$

$1-x^{2}=1-x$

$x^{2}-x=0$

$x(x-1)=0$

$x=1$ atau $x=0$

$V=\pi\int_{0}^{1}\left\{ \left(1-x^{2}\right)^{2}-\left(1-x\right)^{2}\right\} dx$

$=\pi\int_{0}^{1}\left\{ 1-2x^{2}+x^{4}-(1-2x+x^{2}\right\} dx$

$=\pi\int_{0}^{1}\left(x^{4}-3x^{2}+2x\right)dx$

$\begin{aligned}V & =\pi\left[\frac{1}{5}x^{5}-x^{3}+x^{2}\right]_{0}^{1}\\
& =\pi\left(\frac{1}{5}-1+1\right)\\
& =\frac{1}{5}\pi.
\end{aligned}
$

$\begin{aligned}V & =D\\
& =\pi\int_{0}^{\pi}sin^{2}x\cdot dx\\
& =\frac{1}{2}\pi\int_{0}^{\pi}\left(1-cos2x\right)dx
\end{aligned}
$

$\begin{aligned}D & =\frac{1}{2}\pi\left[x-\frac{1}{2}sin2x\right]_{0}^{\pi}\\
& =\frac{1}{2}\pi\left[(\pi-0)-(0-0\right]\\
& =\frac{1}{2}\pi^{2}.
\end{aligned}
$

Jadi nilai $D$ adalah $\frac{1}{2}\pi^{2}.$

Titik potong kurva garis $y=x^{2}$ dan garis $y=2x$

$x^{2}=2x$

$x\left(x-2\right)=0$

$x=0$ atau $x=2$

$\begin{aligned}V & =\pi\int_{0}^{2}\left\{ \left(2x\right)^{2}-\left(x^{2}\right)^{2}\right\} dx\\
& =\pi\int_{0}^{2}\left(4x^{2}-x^{4}\right)dx
\end{aligned}
$

$\begin{aligned}V & =\pi\left[\frac{4}{3}x^{3}-\frac{1}{5}x^{5}\right]_{0}^{2}\\
& =\pi\left(\frac{32}{3}-\frac{32}{5}\right)\\
& =\pi\left(\frac{160-96}{15}\right)\\
& =\frac{64}{15}\pi\\
& =4\frac{4}{15}\pi.
\end{aligned}
$

Cari titik perpotongan antara kurva $y=2x-x^{2}$ dan garis $y=2-x:$

$2x-x^{2}=2-x$

$x^{2}-3x+2=0$

$\left(x-2\right)(x-1)=0$

$x=1$ atau $x=2$

Perhatikan sketsa berikut ini!

$V=\pi\int_{1}^{2}\left\{ \left(2x-x^{2}\right)^{2}-\left(2-x\right)^{2}\right\} dx$

$=\pi\int_{1}^{2}\left(4x^{2}-4x^{3}+x^{4}-(4-4x+x^{2})\right)dx$

$V=\pi\int_{1}^{2}\left(x^{4}-4x^{3}+3x^{2}+4x-4\right)dx$

$=\pi\left[\frac{1}{5}x^{5}-x^{4}+x^{3}+2x^{2}-4x\right]_{1}^{2}$

$V=\pi\left\{ \left(\frac{32}{5}-16+8+8-8\right)-\left(\frac{1}{5}-1+1+2-4\right)\right\} $

$=\pi\left(\frac{32}{5}-16+8+8-8-\frac{1}{5}+2\right)$

$=\pi\left(\frac{31}{5}-6\right)=\frac{1}{5}\pi.$

$y=2\sqrt{1-\frac{x^{2}}{9}}$

$\rightarrow y^{2}=4\left(1-\frac{x^{2}}{9}\right)$

$\begin{aligned}V & =\pi\int_{-3}^{3}y^{2}dx\\
& =2\pi\int_{0}^{3}4\left(1-\frac{x^{2}}{9}\right)dx\\
& =2\pi\int_{0}^{3}\left(4-\frac{4x^{2}}{9}\right)dx
\end{aligned}
$

$\begin{aligned}V & =2\pi\left[4x-\frac{4}{27}x^{3}\right]_{0}^{3}\\
& =2\pi\left(12-\frac{4}{27}\left(3\right)^{3}\right)\\
& =2\pi\left(12-4\right)\\
& =16\pi.
\end{aligned}
$

Kurva $y=7-x^{2}$ terbuka kebawah.

Cari titik perpotongan antara kurva $y=7-x^{2}$ dan $y=x+7:$

$7-x^{2}=x+7$

$x^{2}+x=0$

$x\left(x+1\right)=0$

$x=0$ atau $x=-1$

$V=\pi\int_{-1}^{0}\left\{ \left(7-x^{2}\right)^{2}-\left(x+7\right)^{2}\right\} dx$

$=\pi\int_{-1}^{0}\left(x^{4}-14x^{2}+49-(x^{2}+14x+49)\right)dx$

$V=\pi\int_{-1}^{0}\left(x^{4}-15x^{2}-14x\right)dx$

$=\pi\left[\frac{1}{5}x^{5}-5x^{3}-7x^{2}\right]_{-1}^{0}$

$=\pi\left\{ 0-\left(-\frac{1}{5}+5-7\right)\right\} $

$=\pi\left(2+\frac{1}{5}\right)$

$=\frac{11}{5}\pi.$

Cari titik potong kurva $y=x^{2}$ dan $y=2x-x^{2}$

$x^{2}=2x-x^{2}$

$2x^{2}-2x=0$

$2x(x-1)=0$

$x=0$ atau $x=1$

Perhatikan gambar berikut!

$V=\pi\int_{0}^{1}\left\{ \left(2x-x^{2}\right)^{2}-\left(x^{2}\right)^{2}\right\} dx$

$=\pi\int_{0}^{1}\left(4x^{2}-4x^{3}+x^{4}-x^{4}\right)dx$

$=\pi\int_{0}^{1}\left(4x^{2}-4x^{3}\right)dx$

$\begin{aligned}V & =\pi\left[\frac{4}{3}x^{3}-x^{4}\right]_{0}^{1}\\
& =\pi\left(\frac{4}{3}-1\right)\\
& =\frac{1}{3}\pi.
\end{aligned}
$

$x+y=2$$\rightarrow y=2-x$… (1)

$y=\sqrt{x}$… (2)

Cari titik perpotongan antara kurva $y=\sqrt{x}$ dan garis $x+y=2:$

$2-x=\sqrt{x}$ (kuadratkan kedua ruas)

$x^{2}-4x+4=x$

$x^{2}-5x+4=0$

$\left(x-4\right)\left(x-1\right)=0$

$x=1$ (memenuhi) atau $x=4$ (tidak memenuhi)

Perhatikan gambar berikut!

Untuk mencari volume daerah yang diarsir harus dibagi menjadi dua bagian yaitu volume $D_{1}$ dan volume $D_{2}$

$V=V_{D_{1}}+V_{D_{2}}$

$V=\pi\int_{0}^{1}\left(\sqrt{x}\right)^{2}dx$$+\pi\int_{1}^{2}\left(2-x\right)^{2}dx$

$V=\pi\int_{0}^{1}x.dx$$+\pi\int_{1}^{2}\left(x^{2}-4x+4\right)dx$

$V=\pi\left[\frac{1}{2}x^{2}\right]_{0}^{1}$$+\pi\left[\frac{1}{3}x^{3}-2x^{2}+4x\right]_{1}^{2}$

$V=\pi\left(\frac{1}{2}\right)$$+\pi\left\{ \left(\frac{8}{3}-8+8\right)-\left(\frac{1}{3}-2+4\right)\right\} $

$=\frac{1}{2}\pi+\pi\left(\frac{8}{3}-\frac{1}{3}-2\right)$

$=\frac{1}{2}\pi+\frac{1}{3}\pi=\frac{5}{6}\pi.$

Garis $g$ menyinggung fungsi $y=\sin x$ di titik $(\pi,\,0)$

Gradient garis $g:m$$=\frac{dy}{dx}=cosx$$=cos(\pi)=-1$

Jadi persamaan garis $y=-x+c$ yang melalui titik $(\pi,0):$

$y=-x+\pi$

Perhatikangambar berikut!

$V=\pi\int_{\frac{\pi}{2}}^{\pi}\left\{ \left(-x+\pi\right)^{2}-sin^{2}x\right\} dx$

$=\pi\int_{\frac{\pi}{2}}^{\pi}\left(\left(-x+\pi\right)^{2}-\left(\frac{1}{2}-\frac{1}{2}cos2x\right)\right)dx$

$V=\pi\left[-\frac{1}{3}\left(-x+\pi\right)^{3}-\frac{1}{2}x+\frac{1}{4}sin2x\right]_{\frac{\pi}{2}}^{\pi}$

$=\pi\{\left(-\frac{1}{3}(0)^{3}-\frac{1}{2}\pi+\frac{1}{4}sin2\pi\right)$$-\left(-\frac{1}{3}\left(\frac{\pi}{2}\right)^{3}-\frac{1}{4}\pi+\frac{1}{4}sin\pi\right)\}$

$\begin{aligned}V & =\pi\left(-\frac{1}{2}\pi+\frac{\pi^{3}}{24}+\frac{1}{4}\pi\right)\\
& =\pi\left(\frac{\pi^{3}}{24}-\frac{1}{4}\pi\right)\\
& =\frac{\pi^{2}}{24}\left(\pi^{2}-6\right).
\end{aligned}
$

Perhatikan gambar berikut!

$V_{1}=V_{2}$

$\pi\int_{0}^{a}x^{2}dy=\pi\int_{a}^{4}dy$

$\int_{0}^{a}y\cdot dy=\int_{a}^{4}y\cdot dy$

$\left[\frac{1}{2}y^{2}\right]_{0}^{a}=\left[\frac{1}{2}y^{2}\right]_{a}^{4}$

$\frac{1}{2}a^{2}=\frac{1}{2}(4)^{2}-\frac{1}{2}a^{2}$

$a^{2}=8$

$\begin{aligned}a & =\pm\sqrt{8}\\
& =\pm2\sqrt{2}
\end{aligned}
$

Karena $0<a<4$, maka nilai $a$ yang memenuhi adalah $a=2\sqrt{2}.$