Find volume using a cylindrical shell

[hank]

Just need help on the setup...

Region is enclosed by y = 1/x^3 , x = 1, x = 2, y = 0 and is revolved aorund x = -1.

My setup is this:

r(x) = x + 2
h(x) =1 / x^3

And my answer comes to 5pi / 2

The book says the answer should be 7pi / 4.

Can someone confirm if my setup/answer is wrong? If it is, what should the correct setup be?

 

[soroban]

Hello, hank!

You may kick yourself . . .

Region is enclosed by \(\displaystyle y\,=\,\frac{1}{x^3},\;x\,=\,1,\;x\,=\,2,\; y\,=\,0\:\) and is revolved aorund \(\displaystyle x\,=\,-1\)

My setup is this:

. . \(\displaystyle r(x)\:=\:x\,+\,2\;\) ← here!
. . \(\displaystyle h(x) \:=\:\frac{1}{x^3}\)

The radius is: \(\displaystyle x\,+\,1\)

 

[hank]

Thanks very much for the reply!

However, I don't understand why it's only +1.

If the solid revolves around -1, and the figure starts at 1, doesn't that mean it's 2 units away from -1 and therefore x + 2?

 

[tkhunny]

You are not thinking about it correctly.

1 - (-1) = 2 -- The radius at x = 1
1 - (-1) = 3 -- The radius at x = 2

x - (-1) = x+1

 

[hank]

Ya, I'm going to have to revisit the topic a bit. I bombed my quiz on it today.

Thanks for the help tho.