Nevermind, I got it:
∫[0,1]2pi(x+1)(8-2x-6x^2)dx = 41pi/3
The region in the first quadrant bounded by y=6x^2 , 2x+y=8, and the y-axis is rotated about the line x=-1.
The volume of the resulting solid is:
Please help me set up the integral for this one, I'm not sure how to do it. I've tried ∫[-4/3,1] pi (8-2x-1)^2-(6x^2-1)^2 dx, which I know isn't right but I don't know how to fix it.
3 answers
The graphs intersect at (-4/3,32/3) and (1,6)
We are only interested in x>0, so the volume is
using discs
v = ∫[0,1] π(R^2-r^2) dx
where R=8-2x and r=6x^2
v = ∫[0,1] π((8-2x+1)^2-(6x^2+1)^2) dx = 782π/15
Since we are rotating around the line x = -1, the radius is increased by 1, not reduced.
using shells, we have to divide the region into two parts because of the boundary change at y=6
v = ∫[0,6] 2πrh dy + ∫[6,8] 2πrh dy
where r=y+1 and h=√(y/6) or (8-y)/2
v = ∫[0,6] 2π(y+1)√(y/6) dy + ∫[6,8] 2π(y+1)*(8-y)/2 dy
= 184π/5 + 46π/3 = 782π/15
We are only interested in x>0, so the volume is
using discs
v = ∫[0,1] π(R^2-r^2) dx
where R=8-2x and r=6x^2
v = ∫[0,1] π((8-2x+1)^2-(6x^2+1)^2) dx = 782π/15
Since we are rotating around the line x = -1, the radius is increased by 1, not reduced.
using shells, we have to divide the region into two parts because of the boundary change at y=6
v = ∫[0,6] 2πrh dy + ∫[6,8] 2πrh dy
where r=y+1 and h=√(y/6) or (8-y)/2
v = ∫[0,6] 2π(y+1)√(y/6) dy + ∫[6,8] 2π(y+1)*(8-y)/2 dy
= 184π/5 + 46π/3 = 782π/15
Never mind. I rotated around the line y = -1.
0 answers