Question

Question:, where \({\rm{E}}\) is bounded by the cylinder y² + z² = 9and the planes \({\rm{x = 0, y = 3x}}\), and \({\rm{z = 0}}\) in the first octant.

Short answer

The required answer is \(\frac{{{\rm{27}}}}{{\rm{8}}}\).

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.

Write the integration

Integration Region:

\({\rm{E = }}\left\{ {{\rm{(x,y,z)}}\mid {\rm{0}} \le {\rm{x}} \le \frac{{\rm{1}}}{{\rm{3}}}{\rm{,0}} \le {\rm{y}} \le {\rm{3,0}} \le {\rm{z}} \le \sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} } \right\}\)

Find value of

Assess the integral:

\(\begin{array}{c}\int_{\rm{0}}^{\rm{3}} {\int_{\rm{0}}^{{\rm{y/3}}} {\int_{\rm{0}}^{\sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} } {\rm{z}} } } {\rm{dzdxdy = }}\frac{{\rm{1}}}{{\rm{2}}}\int_{\rm{0}}^{\rm{3}} {\int_{\rm{0}}^{{\rm{y/3}}} {\left( {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} \right)} } {\rm{dxdy}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\int_{\rm{0}}^{\rm{3}} {\frac{{\rm{1}}}{{\rm{3}}}} {\rm{y}}\left( {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} \right){\rm{dy}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{6}}}\int_{\rm{0}}^{\rm{3}} {\left( {{\rm{9y - }}{{\rm{y}}^{\rm{3}}}} \right)} {\rm{dy}}\\{\rm{ = }}\frac{{{\rm{27}}}}{{\rm{8}}}\end{array}\)

Therefore, the required solution is \(\frac{{{\rm{27}}}}{{\rm{8}}}\).