Find the volume of the solid obtained by rotating the region bounded by the curves:
y = x^2 and x = y^2 about the line y = 1.
I received the answer pi/30, using the outer radius of (1 - x^2) and the inner radius of (1 - x^(1/2)). Is this correct?
3 answers
Scratch that, I received the answer of 11pi/30, I made a calculation error that first time. Is this correct?
As usual, you can use shells or discs (washers). I think shells is easier here.
Each shell has the line y=1 as its axis, and height the distance between the curves. We integrate on y, since the shells have thickness dy
v = ∫2πrh dy
where r = (1-y) and h = √y-y^2
v = 2π∫[0,1] (1-y)(√y-y^2) dy
= 11π/30
If you want to use washers, then integrate along x, and we have
v = ∫π(R^2-r^2) dx
where R = 1-x^2 and r=1-√x
v = π∫[0,1] (1-x^2)^2 - (1-√x)^2 dx
= 11π/30
Check these solutions against your work, to see where you went wrong.
Each shell has the line y=1 as its axis, and height the distance between the curves. We integrate on y, since the shells have thickness dy
v = ∫2πrh dy
where r = (1-y) and h = √y-y^2
v = 2π∫[0,1] (1-y)(√y-y^2) dy
= 11π/30
If you want to use washers, then integrate along x, and we have
v = ∫π(R^2-r^2) dx
where R = 1-x^2 and r=1-√x
v = π∫[0,1] (1-x^2)^2 - (1-√x)^2 dx
= 11π/30
Check these solutions against your work, to see where you went wrong.
Thank you!
1 answer