Set up and evaluate an integral to find the volume of a region bounded by the equations z =...

Question:

Set up and evaluate an integral to find the volume of a region bounded by the equations {eq}z = -x^2 - y^2 \enspace and \enspace z = 4y {/eq}

Volume of the Region:

Using the triple integrals formula for the volume of the region in cylindrical coordinates which is {eq}V=\int_{\alpha }^{\beta }\int_{h_{1}(\theta )}^{h_{2}(\theta )}\int_{u_{1}(r\cos \theta ,r\sin \theta )}^{u_{2}(r\cos \theta ,r\sin \theta )}r\:f\left ( r\cos \theta ,r\sin \theta ,z \right )dzdrd\theta {/eq} and the conversion formulas are the following {eq}r^{2}=x^{2}+y^{2},\:x=r\cos \theta ,\:y=r\sin \theta {/eq}.

Answer and Explanation:

First let us find the limits,

{eq}4y=-x^{2}-y^{2} {/eq}

Converting to polar coordinates,

{eq}4r\sin \theta =-r^{2} {/eq}

{eq}r=-4\sin \theta {/eq}

Thus, the limits ate the following

{eq}0\leq \theta \leq 2\pi ,\:0\leq r\leq -4\sin \theta ,\:-r^{2} \leq z\leq 4\sin \theta {/eq}

The volume of the region is,

{eq}V=\int_{0}^{2\pi }\int_{0}^{-4\sin \theta}\int_{-r^{2}}^{4\sin \theta}rdzdrd\theta {/eq}

Integrate with respect to {eq}z {/eq}

{eq}V=\int_{0}^{2\pi }\int_{0}^{-4\sin \theta}\left [ z \right ]^{4\sin \theta}_{-r^{2}}rdrd\theta {/eq}

{eq}V=\int_{0}^{2\pi }\int_{0}^{-4\sin \theta}\left ( 4\sin \theta+r^{2} \right )rdrd\theta {/eq}

Integrate with respect to {eq}r {/eq}

{eq}V=\int_{0}^{2\pi }\left [ \frac{r^4}{4}+2r^2\sin \left(\theta \right) \right ]^{-4\sin \theta}_{0}d\theta {/eq}

{eq}V=\int_{0}^{2\pi }\left ( 64\sin ^4 \theta +32\sin ^3 \theta \right )d\theta {/eq}

Integrate with respect to {eq}\theta {/eq}

{eq}V=\left [ 64\left(-\frac{1}{4}\sin ^3 \theta \cos \theta +\frac{3}{8}\left(\theta -\frac{1}{2}\sin 2\theta \right)\right)+32\left(-\cos \theta +\frac{\cos ^3 \theta }{3}\right) \right ]^{2\pi }_{0} {/eq}

{eq}V=48\pi {/eq}


Learn more about this topic:

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Volumes of Shapes: Definition & Examples

from GMAT Prep: Tutoring Solution

Chapter 11 / Lesson 9
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