# Evaluate the triple integral \iiint_e 3zdV, where E is bounded by the cylinder y^2+z^2=9 and the...

## Question:

Evaluate the triple integral {eq}\iiint_e 3zdV {/eq}, where {eq}E {/eq} is bounded by the cylinder {eq}y^2+z^2=9 {/eq} and the plane {eq}x=0,y=3x {/eq} and {eq}z=0 {/eq} in the first octant.

## Triple Integrals

The volume integral {eq}\displaystyle \iiint_{\mathcal{E}} f(x,y,z) dV, {/eq} where

{eq}\displaystyle \mathcal{E}=\{(x,y,z)|\, a\leq x\leq b, u_1(x)\leq y\leq u_2(x), v_1(x,y)\leq z\leq v_2(x,y)\} {/eq} is evaluated as

{eq}\displaystyle \int_a^b\int_{u_1(x)}^{u_2(x)}\int_{v_1(x,y)}^{v_2(x,y)}f(x,y,z)\, dzdydx. {/eq}

## Answer and Explanation:

To evaluate the integral {eq}\displaystyle \iiint_{\mathcal{E}} 3z\ dV, {/eq} where {eq}\displaystyle \mathcal{E} {/eq} is the solid bounded

above by the cylinder {eq}\displaystyle y^2+z^2=9, {/eq}

bellow by the xy-plane, {eq}\displaystyle z=0, {/eq}

in the back by the yz plane, {eq}\displaystyle x=0, {/eq}

and in front by the palne {eq}\displaystyle y=3x. {/eq}

The solid described above, projected onto the xy-plane is the region bounded

by the y-axis, the line {eq}\displaystyle y=3, \text{ (from the cylinder with radius } 3 ), {/eq} and the line {eq}\displaystyle y=3x, {/eq}

therefore, the solid can be described as

{eq}\displaystyle \mathcal{E}=\{(x,y,z)| \ 0\leq x\leq 1, 3x\leq y\leq 3, 0\leq z\leq \sqrt{9-y^2}\}. {/eq}

So, the integral is calculated as

{eq}\displaystyle \begin{align*}\iiint_{\mathcal{E}} 3z\ dV&=\int_0^1\int_{3x}^3\int_0^{\sqrt{9-y^2}} 3z\ dz dy dx\\ &=\int_0^1\int_{3x}^3\left(\frac{3}{2}z^2\bigg\vert_{z=0}^{z=\sqrt{9-y^2}} \right)\ dy dx\\ &=\frac{3}{2}\int_0^1\int_{3x}^3 \left(9-y^2 \right)\ dy dx\\ &=\frac{3}{2}\int_0^1\left(9y-\frac{1}{3}y^3 \right)\bigg\vert_{y=3x}^{y=3}\ dx\\ &=\frac{3}{2}\int_0^1\left(18-27x+9x^3 \right)\ dx\\ &=\frac{3}{2}\left(18x-\frac{27}{2}x^2+\frac{9}{4}x^4 \right)\bigg\vert_0^1\\ &=\frac{3}{2}\left(18-\frac{27}{2}+\frac{9}{4} \right)=\boxed{\frac{81}{8}}. \end{align*} {/eq}