 # Find the volume of the solid in the first octant bounded by the coordinate plane Jaya Legge 2021-11-02 Answered
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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$$\displaystyle\int\int_{{{R}}}{f{{\left({x},{y}\right)}}}{d}{A}$$
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