# 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

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

2021-02-01
Step 1
Given:
The solid bounded by the graphs of the equations:
z=2−y and $$\displaystyle{z}={4}−{y}^{{2}}$$
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:
$$\displaystyle{z}={2}−{y}\to{z}={4}−{y}^{{2}}$$
The limit for x is:
x=0 to x=3
Now we have to find the limit of y:
We have,
$$\displaystyle{z}={2}-{y}\to{z}={4}-{y}^{{2}}$$
$$\displaystyle\Rightarrow{2}-{y}={4}-{y}^{{2}}$$
$$\displaystyle\Rightarrow{y}^{{2}}-{y}-{2}={0}$$
$$\displaystyle\Rightarrow{y}^{{2}}-{2}{y}+{y}-{2}$$
$$\displaystyle\Rightarrow{y}{\left({y}-{2}\right)}+{1}{\left({y}-{2}\right)}={0}$$
$$\displaystyle\Rightarrow{\left({y}-{2}\right)}{\left({y}+{1}\right)}={0}$$
$$\displaystyle\Rightarrow{\left({y}-{2}\right)}={0}{\quad\text{or}\quad}{\left({y}+{1}\right)}={0}$$
y=2 or y =-1
y=-1 is not possible $$\displaystyle{\left[{y}={0}\right]}$$
Hence we get the limit for y is:
y=0 to y=2
Step 3
$$\displaystyle{V}={\int_{{{x}={0}}}^{{{x}={3}}}}{\int_{{{y}={0}}}^{{{y}={2}}}}{\int_{{{z}={2}-{y}}}^{{{z}={4}-{y}^{{2}}}}}{\left.{d}{z}\right.}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{{x}={0}}}^{{{x}={3}}}}{\int_{{{y}={0}}}^{{{y}={2}}}}{{\left[{z}\right]}_{{{2}-{y}}}^{{{4}-{y}^{{2}}}}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{{x}={0}}}^{{{x}={3}}}}{\int_{{{y}={0}}}^{{{y}={2}}}}{\left[{4}-{y}^{{2}}-{2}+{y}\right]}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{{x}={0}}}^{{{x}={3}}}}{\int_{{{y}={0}}}^{{{y}={2}}}}{\left[{2}-{y}^{{2}}+{y}\right]}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{{x}={0}}}^{{{x}={3}}}}{{\left[{2}{y}-\frac{{y}^{{3}}}{{3}}+\frac{{y}^{{2}}}{{2}}\right]}_{{0}}^{{2}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{{x}={0}}}^{{{x}={3}}}}{\left[{4}-\frac{{8}}{{3}}+{2}\right]}{\left.{d}{x}\right.}$$
$$\displaystyle=\frac{{10}}{{3}}{\int_{{{x}={0}}}^{{{x}={3}}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\frac{{10}}{{3}}{{\left[{x}\right]}_{{0}}^{{3}}}$$
$$\displaystyle=\frac{{10}}{{3}}\times{3}$$
=10
Hence, the volume of the solid bounded by the given graphs of the equations is 10.

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