# Find the volume of the solid in the first octant bounded by the coordinate plane

Find the volume of the solid in the first octant bounded by the coordinate planes, the plane x = 3, and the parabolic cylinder
$$\displaystyle{z}={4}-{y}^{{{2}}}$$.

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Faiza Fuller
$$\displaystyle\int\int_{{{R}}}{f{{\left({x},{y}\right)}}}{d}{A}$$
To find the value of double integral, we must found the area of integration.
It is given:
$$\displaystyle{f{{\left({x},{y}\right)}}}={4}-{y}^{{{2}}}$$
The domain is restricted by the area x=3.
Now, we can solve the integral:
$$\displaystyle{\int_{{{0}}}^{{{2}}}}{\int_{{{0}}}^{{{3}}}}{\left({4}-{y}^{{{2}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={\int_{{{0}}}^{{{2}}}}{{\left[{4}{x}-{y}^{{{2}}}{x}\right]}_{{{0}}}^{{{3}}}}{\left.{d}{y}\right.}$$
$$\displaystyle={\int_{{{0}}}^{{{2}}}}{\left({12}-{3}{y}^{{{2}}}\right)}{\left.{d}{y}\right.}$$ (Use $$\displaystyle\int{x}^{{{a}}}{\left.{d}{x}\right.}={\frac{{{x}^{{{a}+{1}}}}}{{{a}+{1}}}}$$)
$$\displaystyle={{\left[{12}{y}-{y}^{{{3}}}\right]}_{{{0}}}^{{{2}}}}$$
=24-8
=16
Results: 16
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