×
Get Full Access to Calculus: Early Transcendental Functions - 6 Edition - Chapter 14.6 - Problem 4
Get Full Access to Calculus: Early Transcendental Functions - 6 Edition - Chapter 14.6 - Problem 4

×

# ?In Exercises 1 - 8, evaluate the triple iterated integral. $$\int_{0}^{9} \int_{0}^{y / 3} \int_{0}^{\sqrt{y^{2}-9 x^{2}}} z d z d x d y$$

ISBN: 9781285774770 141

## Solution for problem 4 Chapter 14.6

Calculus: Early Transcendental Functions | 6th Edition

• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants

Calculus: Early Transcendental Functions | 6th Edition

4 5 1 402 Reviews
21
3
Problem 4

In Exercises 1 - 8, evaluate the triple iterated integral.

$$\int_{0}^{9} \int_{0}^{y / 3} \int_{0}^{\sqrt{y^{2}-9 x^{2}}} z d z d x d y$$

Text Transcription:

int_{0}^{9} int_{0}^{y / 3} int_{0}^{sqrt{y^{2} - 9x^2}} z dz dx dy

Step-by-Step Solution:

Step 1 of 5) Areas of Surfaces of Revolution In Section 6.4 we found integral formulas for the area of a surface when a curve is revolved about a coordinate axis. Specifically, we found that the surface area is S = 12py ds for revolution about the x-axis, and S = 12px ds for revolution about the y-axis. If the curve is parametrized by the equations x = ƒ(t) and y = g(t), a … t … b, where ƒ and g are continuously differentiable and (ƒ)2 + (g)2 7 0 on 3a, b4 , then the arc length differential ds is given by Equation (4). This observation leads to the following formulas for area of surfaces of revolution for smooth parametrized curves.

Step 2 of 2

##### ISBN: 9781285774770

Since the solution to 4 from 14.6 chapter was answered, more than 239 students have viewed the full step-by-step answer. Calculus: Early Transcendental Functions was written by and is associated to the ISBN: 9781285774770. This full solution covers the following key subjects: . This expansive textbook survival guide covers 134 chapters, and 10738 solutions. The full step-by-step solution to problem: 4 from chapter: 14.6 was answered by , our top Calculus solution expert on 11/14/17, 10:53PM. The answer to “?In Exercises 1 - 8, evaluate the triple iterated integral.$$\int_{0}^{9} \int_{0}^{y / 3} \int_{0}^{\sqrt{y^{2}-9 x^{2}}} z d z d x d y$$Text Transcription:int_{0}^{9} int_{0}^{y / 3} int_{0}^{sqrt{y^{2} - 9x^2}} z dz dx dy” is broken down into a number of easy to follow steps, and 33 words. This textbook survival guide was created for the textbook: Calculus: Early Transcendental Functions, edition: 6.

## Discover and learn what students are asking

Calculus: Early Transcendental Functions : Basic Differentiation Rules and Rates of Change
?Finding a Derivative In Exercises 3–24, use the rules of differentiation to find the derivative of the function. $$y=x^{7}$$

#### Related chapters

Unlock Textbook Solution

?In Exercises 1 - 8, evaluate the triple iterated integral. $$\int_{0}^{9} \int_{0}^{y / 3} \int_{0}^{\sqrt{y^{2}-9 x^{2}}} z d z d x d y$$