Answer :
The volume of the cylindrical coordinates is [tex]\\&\left[\frac{128}{15}+\frac{8 \pi}{3}\right]\end{aligned}$[/tex].
Consider the integral [tex]$\iiint_E(x+y+z) d V, E$[/tex] is the solid in the first octant that lies under the paraboloid [tex]$z=4-x^2-y^2$.[/tex]
Use cylindrical coordinates to evaluate the integral.
Recollect:
The cylindrical coordinates are:
[tex]$x=r \cos \theta, y=r \sin \theta \text { and } z=z$$x^2+y^2=r^2, d x d y=r d r d \theta, d z=d z .$[/tex]
Evaluate the limits of integration as shown below:
[tex]$\begin{aligned}z & =4-\left(x^2+y^2\right) \\& =4-r^2\end{aligned}$[/tex]
Substitute Z=0 in the above equation.
[tex]$\begin{aligned}4-r^2 & =0 \\r^2 & =4 \\r & =2\end{aligned}$[/tex]
The solid lies in the first octant.
Thus, the lower limit of all the integrals is 0 .
The region D is
[tex]$D=\left\{(r, \theta, z) \mid 0 \leq \theta \leq \frac{\pi}{2}, 0 \leq r \leq 2,0 \leq z \leq 4-r^2\right\}$[/tex]
Evaluate the integral as shown below:
[tex]$\begin{aligned}& \iiint_E(x+y+z) d V=\iiint_E(x+y+z) d z d y d x \\& =\int_0^{\frac{\pi}{2}} \int_0^2 \int_0^{4-r^2}(r \cos \theta+r \sin \theta+z) r d z d r d \theta \\& =\int_0^{\frac{\pi}{2}} \int_0^2\left[(r \cos \theta+r \sin \theta) z+\frac{z^2}{2}\right]_0^{4-r^2} r d r d \theta \\& =\int_0^{\frac{\pi}{2}} \int_0^2\left[(r \cos \theta+r \sin \theta)\left(4-r^2\right)+\frac{\left(4-r^2\right)^2}{2}\right] r d r d \theta \\\end[/tex]
[tex]& =\int_0^{\frac{\pi}{2}} \int_0^2\left[(\cos \theta+\sin \theta)\left(4 r^2-r^4\right)+\frac{\left(16 r+r^5-8 r^3\right)}{2}\right] d r d \theta \\& =\int_0^{\frac{\pi}{2}}\left[(\cos \theta+\sin \theta)\left(\frac{4 r^3}{3}-\frac{r^5}{5}\right)+\frac{1}{2}\left(8 r^2+\frac{r^6}{6}-2 r^4\right)\right]_0^2 d \theta \\&\end{aligned}$[/tex]
Integrate w.r.to r
[tex]$\begin{aligned}& \iiint_E(x+y+z) d V=\int_0^{\frac{\pi}{2}}\left[(\cos \theta+\sin \theta)\left(\frac{32}{3}-\frac{32}{5}\right)+\frac{1}{2}\left(32+\frac{64}{6}-32\right)\right] d \theta \\&=\int_0^{\frac{\pi}{2}}\left[\frac{64}{15}(\cos \theta+\sin \theta)+\frac{16}{3}\right] d \theta \quad \text { Simplify. } \\&=\left[\frac{64}{15}(\sin \theta-\cos \theta)+\frac{16}{3} \theta\right]_0^{\frac{\pi}{2}} \\\end[/tex][tex]&=\left[\frac{64}{15}\left(\sin \frac{\pi}{2}-\cos \frac{\pi}{2}\right)+\frac{16}{3} \cdot \frac{\pi}{2}-\frac{64}{15}(\sin 0-\cos 0)+\frac{16}{3} \cdot 0\right] \\&=\left[\frac{64}{15}(1-0)+\frac{8 \pi}{3}\right.\left.-\frac{64}{15}(0-1)\right] \\&=\left[\frac{64}{15}+\frac{8 \pi}{3}+\frac{64}{15}\right] \\&=\left[\frac{128}{15}+\frac{8 \pi}{3}\right]\end{aligned}$[/tex]
Therefore, the volume of the solid is [tex]\\&\left[\frac{128}{15}+\frac{8 \pi}{3}\right]\end{aligned}$[/tex].
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