Question

Question:, where\({\rm{E}}\)is bounded by the parabolic \({\rm{x = 4}}{{\rm{y}}^{\rm{2}}}{\rm{ + 4}}{{\rm{z}}^{\rm{2}}}\)and the plane \({\rm{x = 4}}\).

Short answer

The required answer is \(\frac{{{\rm{16\pi }}}}{{\rm{3}}}\).

Step-by-step solution

Concept Introduction

Triple integrals are the three-dimensional equivalents of double integrals. They're a way to add up an unlimited number of minuscule quantities connected with points in a three-dimensional space.

If the area under the curve from to is the definite integral of a function of one variable, then the double integral is equal to the volume under the surface and above the -plane in the integration region.

Find the value of

Let \({\rm{D = }}\left\{ {{\rm{(y,z)}}\mid {{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}} \le {\rm{1}}} \right\}\)

Then

We can write using polar coordinates.

\(\begin{array}{c}{\rm{ = }}\int_{\rm{0}}^{{\rm{2\pi }}} {\int_{\rm{0}}^{\rm{1}} {\left[ {{\rm{8 - 8}}{{\left( {{{\rm{r}}^{\rm{2}}}} \right)}^{\rm{2}}}} \right]} } {\rm{rdrd\theta }}\\{\rm{ = }}\int_{\rm{0}}^{{\rm{2\pi }}} {\int_{\rm{0}}^{\rm{1}} {\left( {{\rm{8 - 8}}{{\rm{r}}^{\rm{4}}}} \right)} } {\rm{rdrd\theta }}\\{\rm{ = }}\int_{\rm{0}}^{{\rm{2\pi }}} {\int_{\rm{0}}^{\rm{1}} {\rm{8}} } {\rm{r - 8}}{{\rm{r}}^{\rm{5}}}{\rm{drd\theta }}\\{\rm{ = }}\int_{\rm{0}}^{{\rm{2\pi }}} {\left[ {{\rm{4}}{{\rm{r}}^{\rm{2}}}{\rm{ - }}\frac{{\rm{4}}}{{\rm{3}}}{{\rm{r}}^{\rm{6}}}} \right]_{\rm{0}}^{\rm{1}}} {\rm{d\theta }}\\{\rm{ = }}\int_{\rm{0}}^{{\rm{2\pi }}} {\rm{4}} {\rm{ - }}\frac{{\rm{4}}}{{\rm{3}}}{\rm{d\theta }}\\{\rm{ = }}\frac{{\rm{8}}}{{\rm{3}}}\int_{\rm{0}}^{{\rm{2\pi }}} {\rm{d}} {\rm{\theta }}\\{\rm{ = }}\frac{{\rm{8}}}{{\rm{3}}}{\rm{ \times 2\pi }}\\{\rm{ = }}\frac{{{\rm{16\pi }}}}{{\rm{3}}}\end{array}\)

Therefore, the required solution is \(\frac{{{\rm{16\pi }}}}{{\rm{3}}}\).