Compute the volume toz=x^{2},z=1,y=0 and y=2

Harlen Pritchard 2021-09-23 Answered

Compute the volume to
\(\displaystyle{z}={x}^{{{2}}},{z}={1},{y}={0}\) and \(y=2\)

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Expert Answer

Ezra Herbert
Answered 2021-09-24 Author has 17613 answers
Step 1
The volume is computed by the following formula.
\(\displaystyle{V}={\int_{{{x}_{{{1}}}}}^{{{x}_{{{2}}}}}}{\int_{{{y}_{{{1}}}}}^{{{y}_{{{2}}}}}}{\int_{{{z}_{{{1}}}}}^{{{z}_{{{2}}}}}}{\left.{d}{z}\right.}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
Frst, find the limits of integration. Put z=1 in \(\displaystyle{z}={x}^{{{2}}}\) to find the limits of x.
\(\displaystyle{1}={x}^{{{2}}}\)
\(\displaystyle{x}=\pm{1}\)
x=-1,1
Step 2
Now, solve the y and z integrals.
\(\displaystyle{V}={\int_{{-{1}}}^{{{1}}}}{\int_{{{0}}}^{{{2}}}}{\int_{{{x}^{{{2}}}}}^{{{1}}}}{\left.{d}{z}\right.}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{-{1}}}^{{{1}}}}{\int_{{{0}}}^{{{2}}}}{\left({z}\right)}{{\mid}_{{{x}^{{{2}}}}}^{{{1}}}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{-{1}}}^{{{1}}}}{\int_{{{0}}}^{{{2}}}}{\left({1}-{x}^{{{2}}}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{-{1}}}^{{{1}}}}{\left({1}-{x}^{{{2}}}\right)}{y}{{\mid}_{{{0}}}^{{{2}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{-{1}}}^{{{1}}}}{2}{\left({1}-{x}^{{{2}}}\right)}{\left.{d}{x}\right.}\)
Evaluate the x integral to find the volume.
\(\displaystyle{V}={\int_{{-{1}}}^{{{1}}}}{2}{\left({1}-{x}^{{{2}}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle={2}{\left({x}-{\frac{{{x}^{{{3}}}}}{{{3}}}}\right)}{{\mid}_{{-{1}}}^{{{1}}}}\)
\(\displaystyle={2}{\left[{\left({1}-{\frac{{{1}}}{{{3}}}}\right)}-{\left(-{1}+{\frac{{{1}}}{{{3}}}}\right)}\right]}\)
\(\displaystyle={\frac{{{8}}}{{{3}}}}\)
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