# Set up and evaluate a triple integral to find the volume of the solid under the cylinder z =...

## Question:

Set up and evaluate a triple integral to find the volume of the solid under the cylinder {eq}z = \sqrt y {/eq} and bounded by the planes {eq}x=0, z=0, x+y=4 {/eq}

## Volume Integrals:

We can use a triple integral to get the volume of a region, just like we can use a double integral to get an area or a line integral to get a length. We will need to use the provided information to set up an appropriate triple integral, then evaluate it to get the volume.

## Answer and Explanation:

The solid is bounded by {eq}z = \sqrt y {/eq} and {eq}y = 0 {/eq}, so we can use the intersection of these surfaces to get the other bound for {eq}z {/eq}. So we have

{eq}0 \leq z \leq \sqrt y {/eq}

Since {eq}z {/eq} is a function of {eq}y {/eq}, it will be easiest if we integrate {eq}y {/eq} last. With this in mind, we bound {eq}x {/eq} as

{eq}\begin{align*} 0 \leq x \leq 4-y \end{align*} {/eq}

Lastly, we use these bounds to bound {eq}y {/eq} above.

{eq}\begin{align*} 4 - y &= 0 \\ y &= 4 \end{align*} {/eq}

So {eq}y \in [0,4] {/eq}. The volume of the solid is

{eq}\begin{align*} V &= \int_0^4 \int_0^{4-y} \int_0^{\sqrt y} dz\ dx\ dy \\ &= \int_0^4 \int_0^{4-y} \sqrt y\ dx\ dy \\ &= \int_0^4 \sqrt y\ (4-y)\ dy \\ &= \int_0^4 4y^{1/2} - y^{3/2}\ dy \\ &= \left [ 4 \cdot \frac23 y^{3/2} - \frac25 y^{5/2} \right ]_0^4 \\ &= \frac83 (4)^{3/2} - \frac25(4)^{5/2} \\ &= \frac{128}{15} \\ &\approx 8.5333 \end{align*} {/eq}

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from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4