Volume Rot. Region: Find Vol. Bounded by Curve, Y=x^2 & x=y^2, Rotated about Y=1

[tree.lee]

Homework Statement

How do you find the volume of the solid obtained by rotating the region bounded by the given curves about the specified lines?

I keep getting a negative number and I"m becoming so frustrated! 11pi/30 is the answer in the textbook. Thank you so very much in advance!

Homework Equations

y=x^2
x=y^2
about y=1

The Attempt at a Solution

Area:
pi[(1-x^2)^2 - (1-x^(1/2))^2)]

Integral:
pi[1-2x^2+x^4-1+2x^(1/2)-x) between x=0 and x= 1

Antiderivative:
pi[(1/5x)^5-(2/3)x^3 + (2/3)x^(3/2)-(1/2)x^2]

Soln:
pi((2/10)-(5/10)) = -3pi/10??

 

[Mark44]

I edited your post, removing most of the BBcode that you used. Physics Forums rules (https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/) discourage the overuse of special font styles in topics.

Fonts:
When posting a new topic do not use the CAPS lock (all-CAPS), bold, oversized, non-standard, or brightly colored fonts, or any combination thereof. They are hard to read and are considered yelling. When replying in an existing topic it is fine to use CAPS or bold to highlight main points.

 

[Mark44]

Homework Statement

How do you find the volume of the solid obtained by rotating the region bounded by the given curves about the specified lines?

I keep getting a negative number and I"m becoming so frustrated! 11pi/30 is the answer in the textbook. Thank you so very much in advance!

Homework Equations

y=x^2
x=y^2
about y=1

The Attempt at a Solution

Area:
pi[(1-x^2)^2 - (1-x^(1/2))^2)]

Integral:
pi[1-2x^2+x^4-1+2x^(1/2)-x) between x=0 and x= 1

Antiderivative:

pi[(1/5x)^5-(2/3)x^3 + (2/3)x^(3/2)-(1/2)x^2]

You have a mistake in the line above. The coefficient 2/3 in the third term is wrong. I get what you reported to be the correct answer.

Soln:
pi((2/10)-(5/10)) = -3pi/10??