# Thread: a volume integration question

1. ## a volume integration question

The area bounded by the parabola y=2x-x^2, the y-axis and the line y=1 is rotated about the X-axis. Find the volume generated.

Find the volume of the solid generated by the rotating the region bounded by the parabola y=1-x^2 and the lines x=1, y=1 about the X-axis.

For both of these questions I used the volume of revolution formula and integrated each respective equation and calculated it between x=1 and x=0.
For BOTH questions i got (8pi)/15 and for both questions, the answer says (7pi)/15. Could someone please point out where i have gone wrong? i realise that this area is a little unusual so maybe i am not working in the right angle. Please help me if you can. Thank You =)

2. It's almost impossible to point out your mistake unless we see your work. If you type out the integral you got for each volume that would help us help you.

3. okayy sorry ><""

for the first question, my equation was 4x^2 - 4x^3 + x^4 and i integrated that to become 4x^3 - x^4 + (x^5) /5 and then when i calculated it from between x = 1 and x=0 , i got 8/15

for the second question, my equation was 1-2x^2+x^4 and i integrated that to become x-(2x^3)/3+(x^5/5) and again after calculating, it became 8/15

For number 1 it seems that you didn't account for the y=1. If you integrate $\pi \int_{0}^{1}\left(2x-x^2 \right)^2dx$, you will get the volume of the parabola and the x-axis. You are taking a chunk out of this by adding in the boundary of y=1. So the actual volume would be $\pi \int_{0}^{1}\left(2x-x^2 \right)^2-(1)^2dx$