Posted by Charmaine on Saturday, October 27, 2012 at 1:13am.
Major arithmetic mess.
After making quick sketch, it was easy to see that all 3 graphs intersect at (1,6)
the sine curve and the parabola also intersect at (2,0)
and the straight line intersects the parabola at (3.5 , 13.5)
The straight line also intersects the curve at appr .58 and again in the third quadrant, but those two intersection points do not lie in our region
Shade in the region containing the point (2,6).
(This point does not enter any calculations, it was merely given to identify the region we want)
We will have to split our region into two parts, draw a vertical line from (2.0) to the line y=3x+3
Rotation around x-axis:
Part 1. from x = 1 to 2 ...
outer radius = 3x+3
inner radius = 6sin(πx/2)
Volume = π∫(3x+3)^2 - 36sin^2 (πx/2) dx from x=1 to x=2
integral of (3x+3)^2 is not too bad , can be done mentally ..
= (1/3)(1/3)(3x+3)^3 = (1/9)(3x+3)^3
the integral of 6 sin^2 (πx/2) is a bit messier
use cos 2A = 1 - 2sin^2 A
cos πx = 1 - 2sin^2 (πx/2)
3cos πx = 3 - 6sin^2 (πx/2)
6sin^2 (πx/2) = 3 - 3cos πx
and the integral of 6sin^2 (πx/2) is 3x + (3/π)cos πx
so the volume of the first part is
π[ (1/9)(3x+3)^3 + 3x + (3/π)cos πx ] from 1 to 2
= π( 81 + 6 + 3/π - (24 + 3 - 3/π) )
and now for the 2nd part:
Volume = π∫(3x+3)^2 - 36(x-2)^4 dx from x = 2 to 3.5
= π[ (1/9)(3x+3)^3 - (36/5)(x-2)^5 ] form 2 to 3.5
= appr 137.7π
so volume when rotated about x-axis is
appr 197.7π or 621.1
arggghhh, where did you get this question??
rotating about the y-axis is even more fun.
from y = 6 sin (πx/2) we will need x^2 !!!!!!
πx/2 = sin^-1 (y/6)
x = (2/π) sin^-1 (y/6) )
we now have to square that, and then integrate ....
It's a homework problem from my Calculus II class...I tried to attempt and failed miserably. Thank you so much for your help. This sheds a little light so hopefully the rest aren't as bad.
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