Asked by Jessica
Find the volume of the solid obtained by rotating the region bounded by y=x^3, y=1, and the y-axis around the x-axis.
What I did was:
V=ç(0,1)pi(1-x^3)^2 dx
v=pi(x^7/7-x^4/2+x)
and I evaluated it for one since 0 is just going to be 0.
and I got 9pi/14
The answer is wrong, can someone please explain me step by step I really don't know what's wrong.
What I did was:
V=ç(0,1)pi(1-x^3)^2 dx
v=pi(x^7/7-x^4/2+x)
and I evaluated it for one since 0 is just going to be 0.
and I got 9pi/14
The answer is wrong, can someone please explain me step by step I really don't know what's wrong.
Answers
Answered by
Reiny
How about taking the cylinder from 0 to 1 with radius 1 - the region below y = x^3
= π(1^2)(1) - π∫x^3 dx
= π - π[(1/4)x^4 ] from 0 to 1
= π - π(1/4 - 0) = (3/4)π
= π(1^2)(1) - π∫x^3 dx
= π - π[(1/4)x^4 ] from 0 to 1
= π - π(1/4 - 0) = (3/4)π
Answered by
Steve
You interpret the washers wrong. If you have a disc of radius R with a hole in it of radius r, the area is
πR^2 - πr^2, not π(R-r)^2
πR^2 - πr^2, not π(R-r)^2
Answered by
Reiny
forgot to square the radius ....
V = π - π∫(x^3)^2 dx
= π - π[(1/7)x^7 ] from 0 to 1
= π - π(1/7 - 0) = (6/7)π
V = π - π∫(x^3)^2 dx
= π - π[(1/7)x^7 ] from 0 to 1
= π - π(1/7 - 0) = (6/7)π
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