Annette Arroyo

2021-01-31

Use a triple integral to find the volume of the solid bounded by the graphs of the equations. $z=2-y,z=4-{y}^{2},x=0,x=3,y=0$

timbalemX

Step 1
Given:
The solid bounded by the graphs of the equations:
z=2−y and $z=4-{y}^{2}$
x=0 and x=3
y=0
We have to find the volume of the solid bounded by the given graphs of the equations.
Step 2
We know that,
The limits for z is:
$z=2-y\to z=4-{y}^{2}$
The limit for x is:
x=0 to x=3
Now we have to find the limit of y:
We have,
$z=2-y\to z=4-{y}^{2}$
$⇒2-y=4-{y}^{2}$
$⇒{y}^{2}-y-2=0$
$⇒{y}^{2}-2y+y-2$
$⇒y\left(y-2\right)+1\left(y-2\right)=0$
$⇒\left(y-2\right)\left(y+1\right)=0$
$⇒\left(y-2\right)=0\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\left(y+1\right)=0$
y=2 or y =-1
y=-1 is not possible $\left[y=0\right]$
Hence we get the limit for y is:
y=0 to y=2
Step 3
$V={\int }_{x=0}^{x=3}{\int }_{y=0}^{y=2}{\int }_{z=2-y}^{z=4-{y}^{2}}dzdydx$
$={\int }_{x=0}^{x=3}{\int }_{y=0}^{y=2}{\left[z\right]}_{2-y}^{4-{y}^{2}}dydx$
$={\int }_{x=0}^{x=3}{\int }_{y=0}^{y=2}\left[4-{y}^{2}-2+y\right]dydx$
$={\int }_{x=0}^{x=3}{\int }_{y=0}^{y=2}\left[2-{y}^{2}+y\right]dydx$
$={\int }_{x=0}^{x=3}{\left[2y-\frac{{y}^{3}}{3}+\frac{{y}^{2}}{2}\right]}_{0}^{2}dx$
$={\int }_{x=0}^{x=3}\left[4-\frac{8}{3}+2\right]dx$
$=\frac{10}{3}{\int }_{x=0}^{x=3}dx$
$=\frac{10}{3}{\left[x\right]}_{0}^{3}$

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