Posted by **Charmaine** on Saturday, October 27, 2012 at 1:13am.

Let R be the region bounded by y=6sin((pi/2)x), y=6(x-2)^2, y=3x+3 containing the point (2,6). Find the area of R, find the volume of R rotated about the x-axis, find the volume of R rotated about the y-axis, and Suppose R is the base of a shape in which cross-sections perpendicular to the x-axis are squares. Find the volume of this shape.

- Calculus -
**Reiny**, Saturday, October 27, 2012 at 9:23am
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) )

= π(60)

= 60π

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 ....

I YIELD!

- Calculus -
**Anonymous**, Monday, October 29, 2012 at 11:01am
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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