# \int_{0}^{\pi }\int_{0}^{3}\int_{0}^{r^{2}}rdzdrd\theta A) Describe the solid determined by the...

## Question:

{eq}\int_{0}^{\pi }\int_{0}^{3}\int_{0}^{r^{2}}rdzdrd\theta {/eq}

A) Describe the solid determined by the region of integration. (Include a sketch)

B) Evaluate the integral to find the volume of the solid.

## Using Cylindrical Coordinate System To Find Volume:

The volume of any solid bounded by {eq}\displaystyle z=z_1(x,y) {/eq} and {eq}\displaystyle z=z_2(x,y) {/eq} and {eq}\displaystyle y=y_1 {/eq} and {eq}\displaystyle y=y_2 {/eq} and {eq}\displaystyle x=x_1 {/eq} and {eq}\displaystyle x_2 {/eq} can be calculated by setting up a triple integral using cylindrical coordinate systems as follows,

$$\displaystyle \text{ Volume }=\int_{x_1}^{x_2}\int_{y_1}^{y_2}\int_{z_1}^{z_2} \ dz \ dy \ dx$$. If the surfaces and curves bounding the volume are consisting of equations elated to a circle i.e. {eq}\displaystyle x^2+y^2=r^2 {/eq} then the integral may ne converted to cylindrical coordinates by using the substitution, {eq}\displaystyle x=r\cos(\theta) {/eq} and {eq}\displaystyle y=r\sin(\theta) {/eq}. this will convert the given integral to,

$$\displaystyle \text{ Volume }=\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}\int_{z_1}^{z_2} \ r \ dz \ dy \ dx$$

A)

The given integral for volume defines a paraboloid with its apex point at the origin and only the half along the right of the Y-Z plane is defined. Any cross-section taken parallel to the X-Y plane above the plane will be a semi-circle with radius 3 and center along the Z-axis. Also, it will extend from the X-Y plane to the plane {eq}\displaystyle z=9 {/eq} as {eq}\displaystyle z=r^2 {/eq} and the maximum value of {eq}\displaystyle r=3 {/eq}. Considering all this, the sketch of the given volume will be as shown below,

B)

The integration can be evaluated as follows,

{eq}\displaystyle \begin{align} \int_{0}^{\pi }\int_{0}^{3}\int_{0}^{r^{2}} \ r \ dz \ dr \ d\theta &=\int_{0}^{\pi }\int_{0}^{3}r[z]_{0}^{r^{2}}\ dr \ d\theta\\ &=\int_{0}^{\pi }\int_{0}^{3}r[r^{2}-0] \ dr \ d\theta\\ &=\int_{0}^{\pi }\int_{0}^{3}r^3 \ dr \ d\theta\\ &=\left[ \theta\right]_{0}^{\pi}\left[ \frac{r^4}{4}\right]_0^3\\ &=[\pi-0]\left[ \frac{81}{4}-0\right]\\ &=\frac{81\pi}{4} \end{align} {/eq}