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

• tree.lee
In summary, the region bounded by the curves y=x^2 and x=y^2, when rotated about the line y=1, has a volume of 11pi/30. However, there was an error in the solution attempt, specifically in the coefficient 2/3 in the third term of the antiderivative. Once corrected, the correct answer can be obtained.

## 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!

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??

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tree.lee said:

## 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!

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.
tree.lee said:
Soln:
pi((2/10)-(5/10)) = -3pi/10??