# Using geometry, evaluate the double integral. doubleintegral_D sqrt(64 - x^2 - y^2) dA over the...

## Question:

Using geometry, evaluate the double integral.

{eq}\iint_{D} \sqrt{64 - x^{2} - y^{2}}dA \text{ over the circular disk } D: x^{2} + y^{2} \leq 64 {/eq}

## Double Integrals:

In much the same way integrating the difference between two curves gives us the area between them, integrating the difference between two surfaces gives us the volume between them. When the surface is one familiar from geometry, we can use geometry to evaluate the integral.

## Answer and Explanation:

Let' think about what the integral is describing. The top surface is

{eq}\begin{align*} z &= \sqrt(64 - x^2 - y^2) \\ z^2 &= 64 - x^2 - y^2 \\ x^2...

See full answer below.

Become a Study.com member to unlock this answer! Create your account

View this answerLet' think about what the integral is describing. The top surface is

{eq}\begin{align*} z &= \sqrt(64 - x^2 - y^2) \\ z^2 &= 64 - x^2 - y^2 \\ x^2 + y^2 + z^2 &= 8^2 \end{align*} {/eq}

This is a sphere with radius 8. Note that the bottom surface is the {eq}xy {/eq}-plane, i.e. {eq}z = 0 {/eq} (that's why it looks like it's not there). Then the circular disk is the horizontal slice of this sphere through the {eq}xy {/eq}-plane, i.e. a circle with radius 8. So the integral describes the volume of the hemisphere having radius 8. We know from geometry that this is just

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14