How Do You Calculate the Volume of a Solid Rotated Around y = -3?

[olicoh]

Homework Statement

In Problems 11-15, let R be the region bounded by the lines y = −2x – 4, y = 6, and x = −2. Each problem will describe a solid generated by rotating R about an axis. Find the volume of that solid.

14) The solid is generated by rotating R about y = −3.

The Attempt at a Solution

R(x)=6-(-3)
r(x)=(-2x-4)-(-3)
A(x)=pi[(9)²-(-2x-1)²] = pi[81-(4x²+4x+1]
V=(pi) int[-5,-2](-4x²-4x+80)
=pi(-4/3x³-2x²+80x)[-5,-2]

Ok this is where I realize something is wrong because when I plug -5 and -2 into the equation, the number is negative, which is impossible since volume cannot be negative. I keep getting -774pi. Ah!

 

[cragar]

i got <br /> \frac{394\pi}{3}<br />
Your integral looks setup correctly re evaluted your integral.

 

[olicoh]

i got <br /> \frac{394\pi}{3}<br />
Your integral looks setup correctly re evaluted your integral.

? How did you get that? I keep getting a negative number because of that 80x.

 

[cragar]

i just evaluated the integral from -5 to -2
F(-2) - F(-5)
actually i got a different number the second time i made a little mistake, but you should get a positive number

 

[olicoh]

Huh... That's odd. I was plugging it into my graphing calculator and it kept spitting out a negative number. I just plugged it into a calculator online and it got 126. Weird. So I'm assuming the answer is 126pi.

Thanks!

 

[cragar]

ya i got 126pi the second time too.
no problem