Posted by Mia on Wednesday, March 2, 2011 at 8:25pm.
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You got a pair of nasties here.
I will try the first one.
Did you make a sketch?
I got the line and the parabola to intersect at (-1,-2) and (3,6)
since we are rotating about a vertical, the volume has the general appearance of
π[integral] x^2 dy
and we need both equations in term of x
line: x = y/2
parabola : x = ±(y+3)^(1/2)
first consider the region bounded by the line y = 2x and the parabola ABOVE the line from (-1,-2) to (1,-2)
volume of that is ..
π[integral] ((6-y/2)^2 - (6-(y+3)^(1/2))^2 ) dy from -2 to 6
= π[integral] (27 - 7y + y^2/4 - 2(y+3)^(1/2) ) dy
= π { 27y - 7y^2/2 + y^3/12 - (4/3)(y+3)^(3/2) } from -2 to 6
= π(162 - 126 + 18 - 36 - (-54 - 14 - 2/3 - 4/3))
= 88π .... (You better check that arithmetic)
now the part of the parabola from the vertex to the horizontal going through (-1,-2) ....
volume = π[integral] ( (6+(y+3)^(1/2))^2 - (6 - (y+3)^(1/2)^2) dy from -3 to -2
= π[integral] 12(y+3)^(1/2) dy from -3 to -2
= π {16(y+3)^(3/2) } from -3 to -2
= 16π
for a total of 16π + 88π
ok, you better check those calculations. It is so hard to do this kind of math and typing in this kind of setting, and so hard NOT to make some kind of silly error.
try the second one.
Notice you will have symmetry in that one, so use only the part in the first quadrant, then double your answer.
1. Use cylindrical shells to compute the volume of the solid formed by revolving the region bounded by y= 2x and y = x^2 -3 about x= 6.
2. Find the volume of the solid formed by revolving the region bounded by
y= e^x , y= e^-x, and y= 3 about y= 5.
I'm desperately trying to understand this. I hate volume!
First, make a plot of region to be revolved on your calculator. If you do not yet know how to do this, get a friend to show you how. It is a very useful tool. I have a link to a similar plot:
http://img834.imageshack.us/i/1299115553.png/
To calculate using cylindrical shells, you need to cut vertical thin slices of the region and revolve each one around the vertical axis x=6, which coincides with the right side of the above plot.
Let f1(x)=2x, and f2(x)=x^2-3
The intersections of f1(x) and f2(x) are at x=-1 and x=3, or at points (-1,-2), (3,6).
The thin slices have a thickness of dx, at a distance r=6-x from the axis. The height of each slice is therefore h=f1(x)-f2(x). The volume of revolution of each slide is therefore
dV = hdx*2*πr
The total volume, V is
V=∫dV
= ∫2πrhdx from -1 to 3.
Substitute r and h
= ∫2π(6-x)(2x-x^2+3)dx from -1 to 3
Expand the integrand and integrate term by term
= 2π ∫(x^3-8x^2+9x+18)dx
= 2π [(1/4)x^4-(8/3)x^3+(9/2)x^2+18x] [from -1 to 3]
= 320π/3
= 335.1 (approximately)
I will leave #2 for your practice.
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