Question

Question: Use a triple integral to find the volume of the given solid. The solid enclosed by the cylinder \({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = 4}}\)and the planes \({\rm{y = - 1}}\) and \({\rm{y + z = 4}}\).

Short answer

Volume is \({\rm{20\pi }}\).

Step-by-step solution

Define volume

The quantity of three-dimensional space filled by the matter is referred to as volume.

Evaluating volume

Begin by looking for \({\rm{y}}\) limits. It can be seen that minimum \({\rm{y}}\)is \({\rm{y = - 1}}\) and \({\rm{y + z = 4}}\), \({\rm{y = 4 - z}}\)is maximum \({\rm{y}}\). Because \({\rm{z}}\) is likewise constrained in the provided phrase \({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = 4}}\) so \({\rm{z > - 2}}\) the second one is the maximum. As a result,

Now it seems that we need to integrate across \({\rm{D}}\), which we know is defined by \({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = 4}}\) (since we just dealt with y), let's convert everything to polar coordinates.

\(\begin{array}{c}{\rm{ = }}\int_{{\rm{ - 2}}}^{\rm{2}} {\int_{{\rm{ - }}\sqrt {{\rm{4 - }}{{\rm{x}}^{\rm{2}}}} }^{\sqrt {{\rm{4 - }}{{\rm{x}}^{\rm{2}}}} } {{\rm{(5 - z)dzdx}}} } \\{\rm{ = }}\int_{\rm{0}}^{{\rm{2\pi }}} {\int_{\rm{0}}^{\rm{2}} {{\rm{(5 - rsin\theta )rdrd\theta }}} } \\{\rm{ = }}\int_{\rm{0}}^{{\rm{2\pi }}} {\int_{\rm{0}}^{\rm{2}} {{\rm{5r - }}{{\rm{r}}^{\rm{2}}}{\rm{sin\theta drd\theta }}} } \end{array}\)

Now we may complete our double integral. Integrate first with regard to\({\rm{r}}\).

\(\begin{array}{c}{\rm{ = }}\int_{\rm{0}}^{{\rm{2\pi }}} {\left( {\frac{{\rm{5}}}{{\rm{2}}}{{\rm{r}}^{\rm{2}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{3}}}{{\rm{r}}^{\rm{3}}}{\rm{sin\theta }}} \right)} _{\rm{0}}^{\rm{2}}{\rm{d\theta }}\\{\rm{ = }}\int_{\rm{0}}^{{\rm{2\pi }}} {\left( {{\rm{10 - }}\frac{{\rm{8}}}{{\rm{3}}}{\rm{sin\theta - 0}}} \right)} {\rm{d\theta }}\end{array}\)

Integrate in the direction of\({\rm{\theta }}\):

\({\rm{ = }}\left( {{\rm{10\theta + }}\frac{{\rm{8}}}{{\rm{3}}}{\rm{cos\theta }}} \right)_{\rm{0}}^{{\rm{2\pi }}}\)

Thus, the result is,

\(\begin{array}{c}{\rm{ = }}\left( {{\rm{20\pi + }}\frac{{\rm{8}}}{{\rm{3}}}{\rm{ - 0 - }}\frac{{\rm{8}}}{{\rm{3}}}} \right)\\{\rm{ = 20\pi }}\end{array}\)

Therefore, the volume is \({\rm{20\pi }}\).