First let's imagine how the solid looks like.
Let's rearrange the equation of the ellipsoid to find semi-axes:
4x² + 4y² + z² = 64 | : 64
x²/16 + y²/16 + z²/64 = 1
So semi-axes are a=4, b=4, c=8. So x varies (-4, 4), y from(-4,4) and z from (-8, 8). So ellipsoid is much wider than the cylinder.
Cylinder defines region in xy plane and the ellipsoid bounds solid from bellow and from above.
Let's find the region in the xy plane. (You may skip this part if you understand from above that the region in xy plane is a disk x² + y² ? 4)
The region R is bounded by the cylinder x² + y² = 4 (1) and projection of the ellipsoid in the xy plane:
z=0 and 4x² + 4y² + z² = 64
4x² + 4y² = 64
x² + y² = 16...(2)
So the region in xy plane is bounded by two circles (1) and (2) and actually is inside both of them:
http://i633.photobucket.com/albums/uu51/labasrasa/Problems-3/circle0415.pngIntroduce polar coordinates
x = r*cost, y = r*sint
Then integration limits for thr region R:
0<= t <= 2pi
0 <= r <= 2
Next let's find the limits for z from the equation of the ellipsoid:
4x² + 4y² + z² = 64
4(x² + y²) + z² = 64
4r² + z² = 64
z² = 64 - 4r²
z² = 4 (16 - r²)
z = +/- 2 ?(16 - r²)
So the integration limits for z : -2 ?(16 - r²) <= z <= 2 ?(16 - r²).
But as the solid is symmetrical we can take just a half and double the volume:
V = 2* (0, 2pi) ? (0, 2)? (0, 2 ?[16 - r²]) ? r dz dr dt
I hope you are able to finish it. If not - just give me a message.
After integration the volume is equal to
V = 512pi/3 - 64pi*?3. <===ANSWER
Hope this helps
Quick Reply
