geduiwelh

2021-09-17

Compute the volume to
$z={y}^{2},z=1,2x+z=4,x=0$

Mayme

Step 1
The volume is computed by the following formula.
$V={\int }_{{x}_{1}}^{{x}_{2}}{\int }_{{y}_{1}}^{{y}_{2}}{\int }_{{z}_{1}}^{{z}_{2}}dzdydx$
First, find the limits of integration. Put z=1 in $z={y}^{2}$ to find the limits of y. Also, find the limits of x.
$1={y}^{2}$
$y=±1$
y=-1,1
2x+z=4
$x=2-\frac{z}{2}$
Step 2
Now, solve the x and z integrals.
$V={\int }_{-1}^{1}{\int }_{{y}^{2}}^{1}{\int }_{0}^{2-\frac{z}{2}}dxdzdy$
$={\int }_{-1}^{1}{\int }_{{y}^{2}}^{1}\left(2-\frac{z}{2}\right)dzdy$
$={\int }_{-1}^{1}\left(2z-\frac{{z}^{2}}{2\cdot 2}\right){\mid }_{{y}^{2}}^{1}dy$
$={\int }_{-1}^{1}\left[\left(2-\frac{1}{4}\right)-\left(2{y}^{2}-\frac{{y}^{4}}{4}\right)\right]dy$
$={\int }_{-1}^{1}\left(\frac{7}{4}-2{y}^{2}+\frac{{y}^{4}}{4}\right)dy$
Evaluate the y integral to find the volume.
$V={\int }_{-1}^{1}\left(\frac{7}{4}-2{y}^{2}+\frac{{y}^{4}}{4}\right)dy$
$=\left(\frac{7y}{4}-\frac{2{y}^{3}}{3}+\frac{{y}^{5}}{20}\right){\mid }_{-1}^{1}$
$=2\left(\frac{7}{4}-\frac{2}{3}+\frac{1}{20}\right)$
$=\frac{34}{15}$

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