Question

Find the volume of the solid that lies between the paraboloid \({\rm{z = }}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}\)and the sphere \({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = 2}}\).

Short answer

The volume of the solid is \(\left( {\frac{{{\rm{8}}\sqrt {\rm{2}} {\rm{ - 7}}}}{{\rm{6}}}} \right){\rm{\pi }}\).

Step-by-step solution

Find the volume of the solid.    

Paraboloid: \({\rm{z = }}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}\) or \({\rm{z = }}{{\rm{r}}^{\rm{2}}}\)

Sphere: \({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = 2}}\) or \({\rm{z = }}\sqrt {{\rm{2 - }}{{\rm{r}}^{\rm{2}}}} \)

NOTE: The bottom bound is paraboloid, while the upper bound is the top section of the sphere.

When it comes to integration, \({{\rm{r}}^{\rm{2}}}{\rm{ = }}\sqrt {{\rm{2 - }}{{\rm{r}}^{\rm{2}}}} \)

Solving for \({\rm{r}}\) gives us \({\rm{r = 1}}\). This is a circle with a radius of one.

So, \({\rm{0}} \le {\rm{r}} \le {\rm{1}}\)and\(0 \le \theta \le 2\pi \)

\(\int_{\rm{0}}^{{\rm{2\pi }}} {\int_{\rm{0}}^{\rm{1}} {\int_{{{\rm{r}}^{\rm{2}}}}^{\sqrt {{\rm{2 - }}{{\rm{r}}^{\rm{2}}}} } {\rm{r}} } } {\rm{dzdrd\theta }}\)

Let's start with \(dz\):

\(\begin{aligned}\int_{{{\rm{r}}^{\rm{2}}}}^{\sqrt {{\rm{2 - }}{{\rm{r}}^{\rm{2}}}} } {\rm{r}} \rm dz &= r\left( {\sqrt {{\rm{2 - }}{{\rm{r}}^{\rm{2}}}} {\rm{ - }}{{\rm{r}}^{\rm{2}}}} \right)\\\rm &= r\sqrt {{\rm{2 - }}{{\rm{r}}^{\rm{2}}}} {\rm{ - }}{{\rm{r}}^{\rm{3}}}\end{aligned}\)

Step 2:Evaluate the equation.

Now, \({\rm{dr}}\)

\(\int_{\rm{0}}^{\rm{1}} {\rm{r}} \sqrt {{\rm{2 - }}{{\rm{r}}^{\rm{2}}}} {\rm{ - }}{{\rm{r}}^{\rm{3}}}{\rm{dr = }}\int_{\rm{0}}^{\rm{1}} {\rm{r}} \sqrt {{\rm{2 - }}{{\rm{r}}^{\rm{2}}}} {\rm{dr - }}\int_{\rm{0}}^{\rm{1}} {{{\rm{r}}^{\rm{3}}}} {\rm{dr}}\)

Let's start with \({\rm{(1) - (2):}}\)

\(\begin{aligned} &\Rightarrow \rm (l) = \int_{\rm{0}}^{\rm{1}} {\rm{r}} \sqrt {{\rm{2 - }}{{\rm{r}}^{\rm{2}}}} {\rm{dr}}\\ &\Rightarrow \rm (2) = \int_{\rm{0}}^{\rm{1}} {{{\rm{r}}^{\rm{3}}}} {\rm{dr}}\end{aligned}\)

Use a u-substitution for (l). \(^*\) Don't forget to adjust the boundaries if necessary. \({\rm{du = - 2rdr}}\) and\({\rm{u = 2 - }}{{\rm{r}}^{\rm{2}}}\).

Note: \( - \int_{\rm{b}}^{\rm{a}} {\rm{ = }} \int_a^b {} \)

\(\begin{aligned}{\rm{ - }}\int_{\rm{2}}^{\rm{1}} {\frac{{\sqrt {\rm{u}} }}{{\rm{2}}}} \rm du &= \int_{\rm{1}}^{\rm{2}} {\frac{{\sqrt {\rm{u}} }}{{\rm{2}}}} {\rm{du}}\\ \rm &= \left( {\frac{{{{\rm{u}}^{{\rm{3/2}}}}}}{{\rm{3}}}} \right)_{\rm{1}}^{\rm{2}}\\ \rm &= \frac{{{\rm{2}}\sqrt {\rm{2}} {\rm{ - 1}}}}{{\rm{3}}}\end{aligned}\)

Now, \({\rm{(2)}}\) is a small thing.

\(\begin{aligned}\int_{\rm{0}}^{\rm{1}} {{{\rm{r}}^{\rm{3}}}} \rm dr &= \frac{{\rm{1}}}{{\rm{4}}}\\\rm (1) - (2) &= \frac{{{\rm{8}}\sqrt {\rm{2}} {\rm{ - 7}}}}{{{\rm{12}}}}\end{aligned}\)

Let's finish with \({\rm{d\theta }}\).

\(\begin{aligned}\int_{\rm{0}}^{{\rm{2\pi }}} {\frac{{{\rm{8}}\sqrt {\rm{2}} {\rm{ - 7}}}}{{{\rm{12}}}}} \rm d\theta &= \left( {\frac{{{\rm{8}}\sqrt {\rm{2}} {\rm{ - 7}}}}{{{\rm{12}}}}} \right){\rm{(\theta )}}_{\rm{0}}^{{\rm{2\pi }}}\\\rm &= \left( {\frac{{{\rm{8}}\sqrt {\rm{2}} {\rm{ - 7}}}}{{{\rm{12}}}}} \right){\rm{2\pi }}\\\rm &= \left( {\frac{{{\rm{8}}\sqrt {\rm{2}} {\rm{ - 7}}}}{{\rm{6}}}} \right){\rm{\pi }}\end{aligned}\)

Therefore, the volume of the solid is \(\left( {\frac{{{\rm{8}}\sqrt {\rm{2}} {\rm{ - 7}}}}{{\rm{6}}}} \right){\rm{\pi }}\).