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.

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.