the curves intersect at (±2,√8), and the vertex is at (0,2)
the area is symmetric about the y-axis.
So, using shells of thickness dy, we have
v = 2∫[2,√8] 2πrh dy
where r = y and h = x
v = 2∫[2,√8] 2πy√(y^2-4) dy = 32π/3
using discs (washers) of thickness dx,
v = 2∫[0,4] π(R^2-r^2) dx
where R = 3 and r = y
2∫[0,4] π(9-(x^2+4)) dx = 44π/3
Hmmm. I have to leave just now -- maybe you can spot my mistake.
I'll be back in a few hours, in case it's still a mystery.
The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method.
y2 − x2 = 4, y = 3; about the x-axis
4 answers
that should be
2∫[0,4] π(8-(x^2+4)) dx = 32π/3
2∫[0,4] π(8-(x^2+4)) dx = 32π/3
dang! I used the wrong limit for x.
If you see it, fix it and redo the math
If you see it, fix it and redo the math
okay - I've calmed down now. Here's the real solution.
the curves intersect at (±√5,3), and the vertex is at (0,2)
the area is symmetric about the y-axis.
So, using shells of thickness dy, we have
v = 2∫[2,3] 2πrh dy
where r = y and h = x
v = 2∫[2,3] 2πy√(y^2-4) dy = (20π√5)/3
using discs (washers) of thickness dx,
v = 2∫[0,√5] π(R^2-r^2) dx
where R = 3 and r = y
2∫[0,√5] π(9-(x^2+4)) dx = (20π√5)/3
the curves intersect at (±√5,3), and the vertex is at (0,2)
the area is symmetric about the y-axis.
So, using shells of thickness dy, we have
v = 2∫[2,3] 2πrh dy
where r = y and h = x
v = 2∫[2,3] 2πy√(y^2-4) dy = (20π√5)/3
using discs (washers) of thickness dx,
v = 2∫[0,√5] π(R^2-r^2) dx
where R = 3 and r = y
2∫[0,√5] π(9-(x^2+4)) dx = (20π√5)/3
2 answers