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

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
asked 2021-01-31
Use a triple integral to find the volume of the solid bounded by the graphs of the equations. \(\displaystyle{z}={2}−{y},{z}={4}−{y}^{{2}},{x}={0},{x}={3},{y}={0}\)

Answers (1)

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.
0

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