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

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

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Ezra Herbert
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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