# Use the triple integral to find the volume of the given solid. The solid enclosed by the...

## Question:

Use the triple integral to find the volume of the given solid. The solid enclosed by the paraboloid {eq}x = 5y^2 + 5z^2 {/eq} and the plane {eq}x = 14. {/eq}

## Finding the Volume:

Let us find the volume of the given solid in the triple integral form by converting the given integral function into polar coordinates.

The general form of the volume in polar coordinate is {eq}V = \iiint_{E} f \left(x, r, \theta \right) \ dV {/eq}. Where, {eq}dV = dx \ dr \ d\theta {/eq}.

When equating the given equation we will get the limit for {eq}r \ and \ x {/eq}. By polar coordinate the {eq}\theta {/eq} range varies from {eq}0 \ to \ 2\pi {/eq}.

## Answer and Explanation:

The given paraboloid equation is:

{eq}x = 5y^2 + 5z^2 \quad equation \cdots \left(1 \right) {/eq}

The given plane equation is:

{eq}x = 14 \quad equation \cdots \left(2 \right) {/eq}

To find volume:

{eq}V = \iiint_{E} f \left(x, r, \theta \right) \ dV {/eq}

By using Polar coordinates:

{eq}y = r\cos \theta, z = r\sin \theta, x = x \\ y^{2} + z^{2} = \left ( r\cos \theta \right )^{2} + \left ( r\sin \theta \right )^{2} \\ y^{2} + z^{2} = r^{2} \left ( \cos^{2} \theta + \sin^{2} \theta \right ) \\ y^{2} + z^{2} = r^{2} {/eq}

Let us find the limits for integration:

{eq}x = 5y^2 + 5z^2 \\ x = 5 \left(y^2 + z^2 \right) \\ x = 5r^2 {/eq}

Let's equating the equations (1) and (2):

{eq}x = x \\ 5y^2 + 5z^2 = 14 \\ 5 \left(y^2 + z^2 \right) = 14 \\ y^2 + z^2 = \frac{14}{5} \\ r^2 = \frac{14}{5} \\ r = \sqrt{\frac{14}{5}} {/eq}

We know that, the {eq}\theta {/eq} range varies from {eq}0 \ to \ 2\pi {/eq}.

The limits are:

{eq}5r^2 \leq x \leq 14 \\ 0 \leq r \leq \sqrt{\frac{14}{5}} \\ 0 \leq \theta \leq 2\pi {/eq}

Now, we are going to integrate the triple integral function to find the volume by using polar coordinates:

{eq}\begin{align*} \displaystyle V &= \iiint_{E} f \left[ x, r, \theta \right] \ dV \\ &= \int_{0}^{2\pi} \int_{0}^{\sqrt{\frac{14}{5}}} \int_{5r^2}^{14} \ r \ dx \ dr \ d\theta \\ &= \int_{0}^{2\pi} \int_{0}^{\sqrt{\frac{14}{5}}} \left [ xr \right ]_{5r^2}^{14} \ dr \ d\theta \\ &= \int_{0}^{2\pi} \int_{0}^{\sqrt{\frac{14}{5}}} \left [ 14 \cdot r - 5r^2 \cdot r \right ] \ dr \ d\theta \\ &= \int_{0}^{2\pi} \int_{0}^{\sqrt{\frac{14}{5}}} \left [ 14r - 5r^3\right ] \ dr \ d\theta \\ &= \int_{0}^{2\pi} \left [ 7r^2 - \frac{5r^4}{4} \right ]_{0}^{\sqrt{\frac{14}{5}}} \ d\theta \\ &= \int_{0}^{2\pi} \left [ 7 \left ( \sqrt{\frac{14}{5}} \right )^2 - \frac{5\left ( \sqrt{\frac{14}{5}} \right )^4}{4} - 0 \right ] \ d\theta \\ &= \int_{0}^{2\pi} \left [ \frac{98}{5} - \frac{49}{5} \right ] \ d\theta \\ &= \int_{0}^{2\pi} \left [\frac{49}{5} \right ] \ d\theta \\ &= \left [ \frac{49}{5} \theta \right ]_{0}^{2\pi} \\ &= \left [ \frac{49}{5} \cdot 2\pi - 0 \right ] \\ &= \left [ \frac{98\pi }{5} \right ] \\ \displaystyle \therefore V &= 61.57522 \end{align*} {/eq}

Therefore, the volume is {eq}\displaystyle 61.57522 \ cubic \ units {/eq}.

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from CAHSEE Math Exam: Help and Review

Chapter 18 / Lesson 12