- #1

jaredm2012

- 5

- 0

Nevermind, I finally figured it out!

Find the volume of the solid generated by revolving the region bounded by y=x^1/2 and the lines y=2 and x=0 about:

a. the x-axis

b. the y-axis

c. the line y=2

integral from a--b pi[R(x)]^2dx

integral from a--b pi([R(x)]^2 - [r(x)]^2)dx

I solved a, b, and c just fine.

a = 8pi

b = 32pi/5

c = 8pi/3

However I am confused about part d, particularly how to find the radius. The book gives the answer as 224pi/15, but I get 256pi/15.

I did the following:

pi*integral from 0 to 2 (4-y^2)^2dy, which gave me 256pi/15 when I worked it out. I may just be messing up the math, though I did the problem a few times the same way and got the same answer every time. I also tried using pi*integral from 0 to 2 16-y^4, but this also yielded the wrong answer. Basically I am not sure how to find the radius of the region it is asking for, when rotating around the line x=4.

## Homework Statement

Find the volume of the solid generated by revolving the region bounded by y=x^1/2 and the lines y=2 and x=0 about:

a. the x-axis

b. the y-axis

c. the line y=2

__d. the line x=4__## Homework Equations

integral from a--b pi[R(x)]^2dx

integral from a--b pi([R(x)]^2 - [r(x)]^2)dx

## The Attempt at a Solution

I solved a, b, and c just fine.

a = 8pi

b = 32pi/5

c = 8pi/3

However I am confused about part d, particularly how to find the radius. The book gives the answer as 224pi/15, but I get 256pi/15.

I did the following:

pi*integral from 0 to 2 (4-y^2)^2dy, which gave me 256pi/15 when I worked it out. I may just be messing up the math, though I did the problem a few times the same way and got the same answer every time. I also tried using pi*integral from 0 to 2 16-y^4, but this also yielded the wrong answer. Basically I am not sure how to find the radius of the region it is asking for, when rotating around the line x=4.

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