$\begin{aligned}V & =\pi\int_{0}^{3}y^{2}dy\\
& =\pi\left[\frac{1}{3}y^{3}\right]_{0}^{3}\\
& =9\pi.
\end{aligned}
$

$y=4-x^{2}$

$\rightarrow x^{2}=4-y$

Perpotongan terhadap sumbu $y$ jika $x=0$

$f(0)=4-0^{2}=0$

$\rightarrow y=4$

$\begin{aligned}V & =\pi\int_{0}^{4}x^{2}dy\\
& =\pi\int_{0}^{4}(4-y)dy\\
& =\pi\left[4y-\frac{1}{2}y^{2}\right]_{0}^{4}\\
& =\pi\left(16-8\right)\\
& =8\pi.
\end{aligned}
$

Perhatikan gambar berikut!

$\begin{aligned}V & =\pi\int_{0}^{4}x^{2}dy\\
& =\pi\int_{0}^{4}y\cdot dy\\
& =\pi\left[\frac{1}{2}y^{2}\right]_{0}^{4}\\
& =8\pi.
\end{aligned}
$

$xy=1$

$\rightarrow x=\frac{1}{y}$

$\leftrightarrow x^{2}=\frac{1}{y^{2}}$

$\begin{aligned}V & =\pi\int_{2}^{3}x^{2}dy\\
& =\pi\int_{2}^{3}\frac{1}{y^{2}}\cdot dy\\
& =\pi\left[-\frac{1}{y}\right]_{2}^{3}\\
& =\pi\left(-\frac{1}{3}-\left(-\frac{1}{2}\right)\right)\\
& =\pi\left(\frac{1}{2}-\frac{1}{3}\right)\\
& =\frac{1}{6}\pi.
\end{aligned}
$

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

$\rightarrow x^{2}=4-y$

$\begin{aligned}V & =\pi\int_{-1}^{0}x^{2}\cdot dy\\
& =\pi\int_{-1}^{0}\left(4-y\right)dy\\
& =\pi\left[4y-\frac{1}{2}y^{2}\right]_{-1}^{0}\\
& =\pi\left(0-\left(-4-\frac{1}{2}\right)\right)\\
& =4\frac{1}{2}\pi\\
& =\frac{9}{2}\pi.
\end{aligned}
$

$y=\sqrt{x}\rightarrow x^{2}=y^{4}$

Perhatikan gambar berikut!

$\begin{aligned}V & =\pi\int_{0}^{2}\left(4^{2}-y^{4}\right)dy\\
& =\pi\int_{0}^{2}\left(16-y^{4}\right)dy\\
& =\pi\left[16y-\frac{1}{5}y^{2}\right]_{0}^{2}\\
& =\pi\left(32-\frac{32}{5}\right)\\
& =\frac{128}{5}\pi.
\end{aligned}
$

Perhatikan kurva berikut :

$\begin{aligned}V & =2\pi\int_{2}^{6}\left(y-2\right)dy\\
& =2\pi\left[\frac{1}{2}y^{2}-2y\right]_{2}^{6}\\
& =2\pi\left\{ \left(18-12\right)-\left(2-4\right)\right\} \\
& =2\pi\left(6+2\right)\\
& =16\pi.
\end{aligned}

Cari titik perpotonga dari kurva $y=-x^{2}+4$ dan $y=-2x+4$

$\begin{aligned}-x^{2}+4 & =-2x+4\\
x^{2}-2x & =0\\
x\left(x-2\right) & =0
\end{aligned}
$

$x=0$ atau $x=2$

untuk $x=0$, maka diperoleh nilai $y=4$

untuk nilai $x=2$, maka diperoleh $y=0$

$y=-x^{2}+4$$\rightarrow x^{2}=4-y$

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

$\begin{aligned}V & =\pi\int_{0}^{4}\left\{ \left(4-y\right)-\left(\frac{4-y}{2}\right)^{2}\right\} dy\\
& =\pi\int_{0}^{4}\left(4-y-\left(\frac{16-8y+y^{2}}{4}\right)\right)dy
\end{aligned}
$

$\begin{aligned}V & =\pi\int_{0}^{4}\left(\frac{16-4y-16+8y-y^{2}}{4}\right)dy\\
& =\frac{\pi}{4}\int_{0}^{4}\left(-y^{2}+4y\right)\\
& =\frac{\pi}{4}\left[-\frac{1}{3}y^{3}+2y^{2}\right]_{0}^{4}\\
& =\frac{\pi}{4}\left(-\frac{64}{3}+32\right)\\
& =\frac{\pi}{4}\left(\frac{96-64}{3}\right)\\
& =\frac{32}{12}\pi\\
& =\frac{8}{3}\pi.
\end{aligned}
$

$y=4x^{2}$$\rightarrow x^{2}=\frac{y}{4}$

Perhatikan gambar berikut!

$\begin{aligned}V & =\pi\int_{0}^{4}x^{2}dy\\
& =\pi\int_{0}^{4}\frac{y}{4}\cdot dy\\
& =\pi\left[\frac{1}{8}y^{2}\right]_{0}^{4}\\
& =2\pi.
\end{aligned}
$

$y=\sqrt{x}\rightarrow x^{2}=y^{4}$

Untuk $y=2$, diperoleh nilai $x=4$

$\begin{aligned}V & =\pi\int_{0}^{2}\left(4^{2}-y^{4}\right)dy\\
& =\pi\int_{0}^{2}\left(16-y^{4}\right)dy\\
& =\pi\left[16y-\frac{1}{5}y^{5}\right]_{0}^{2}\\
& =\pi\left(32-\frac{32}{5}\right)\\
& =\frac{128}{5}\pi\\
& =25\frac{3}{5}\pi.
\end{aligned}
$

$x+y=4$$\rightarrow x=4-y$ …. (1)

$x=\left(y-2\right)^{2}$…. (2)

Cari titik perpotongan antara kurva $x=\left(y-2\right)^{2}$ dan garis $x+y=4$ dengan cara mensubtitusikan pers (1) ke pers (2) :

$\left(y-2\right)^{2}=4-y$

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

$y^{2}-3y=0$

$y\left(y-3\right)=$0

$y=0$ atau $y=3$

Perhatikan gambar berikut!

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

$\begin{aligned}V & =\pi\left[-\frac{1}{3}\left(4-y\right)^{3}-\frac{1}{5}\left(y-2\right)^{5}\right]_{0}^{3}\\
& =\pi\left[\left(-\frac{1}{3}-\frac{1}{5}\right)-\left(-\frac{64}{3}+\frac{32}{5}\right)\right]\\
& =\pi\left(\frac{63}{3}-\frac{33}{5}\right)\\
& =\frac{216}{15}\pi\\
& =\frac{72}{5}\pi.
\end{aligned}
$

$x=\sqrt{y}\rightarrow x^{2}=y$

$y=4x^{2}\rightarrow x^{2}=\frac{1}{4}y$

$\begin{aligned}V & =\pi\int_{0}^{4}\left(y-\frac{1}{4}y\right)dy\\
& =\pi\int_{0}^{4}\frac{3}{4}y\cdot dy\\
& =\pi\left[\frac{3}{8}y^{2}\right]_{0}^{4}\\
& =6\pi.
\end{aligned}
$

Cari titik perpotongan antara kurva $y=x^{2}$ dan garis $y=3x$

$x^{2}=3x$

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

$x\left(x-3\right)=0$

$x=0$ atau $x=3$

untuk $x=0$ diperoleh $y=0$

untuk $x=3$ diperoleh $y=9$

Perhatikan gambar berikut!

$\begin{aligned}V & =\pi\int_{0}^{9}\left\{ y-\left(\frac{y}{3}\right)^{2}\right\} dy\\
& =\pi\int_{0}^{9}\left(y-\frac{y^{2}}{9}\right)dy\\
& =\pi\left[\frac{1}{2}y^{2}-\frac{1}{27}y^{3}\right]_{0}^{9}\\
& =\pi\left(\frac{81}{2}-27\right)\\
& =\frac{27}{2}\\
& =13\frac{1}{2}\pi.
\end{aligned}
$

Perhatikan gambar berikut!

$V_{1}=V_{2}$

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

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

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

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

$a^{2}=8$

$a=\pm\sqrt{8}=\pm2\sqrt{2}$

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

$*\ x=2\sqrt{2}y^{2}$$\rightarrow x^{2}=8y^{4}$…. (1)

$*\ x^{2}+y^{2}=9$…. (2)

Cari titik perptongan kedua kurva dengan cara mensubstitusikan pers (1) ke pers (2) :

$8y^{4}+y^{2}=9$

$8y^{4}+y^{2}-9=0$

$\left(8y^{2}+9\right)\left(y^{2}-1\right)=0$

$y^{2}=-\frac{9}{8}\rightarrow y=\sqrt{-\frac{9}{8}}$ ( tidak memenuhi)

$y^{2}=1\rightarrow y=\pm1$

Perhatikan gambar berikut!

$\begin{aligned}V_{1} & =\pi\int_{1}^{3}(9-y^{2})dy\\
& =\pi\left[9y-\frac{1}{3}y^{3}\right]_{1}^{3}\\
& =\pi\left\{ \left(27-9\right)-\left(9-\frac{1}{3}\right)\right\} \\
& =9\frac{1}{3}\pi
\end{aligned}
$

$\begin{aligned}V_{2} & =\pi\int_{0}^{1}8y^{4}\cdot dy\\
& =\pi\left[\frac{8}{5}y^{5}\right]_{0}^{1}\\
& =\frac{8}{5}\pi
\end{aligned}
$

Volume daerah yang diarsir

$\begin{aligned}V & =V_{1}+V_{2}\\
& =9\frac{1}{3}\pi+\frac{8}{5}\pi\\
& =\frac{28}{3}\pi+\frac{8}{5}\pi\\
& =\frac{164}{15}\pi.
\end{aligned}
$