# Thread: Volume using Shell method. Stuck on last step!

1. ## Volume using Shell method. Stuck on last step!

The volume of this solid can also be computed using cylindrical shells via an integral
2pi(x+4)(4x-x^2) dx

with limits of integration 0 and 4

volume is cubic units.

I am stuck on the last step. For some reason I tried integrating it and then plugging in the value, which in this case is just the beta, 4.

What am I doing wrong? this is supposed to be easy step..

2. Originally Posted by softballchick
The volume of this solid can also be computed using cylindrical shells via an integral
2pi(x+4)(4x-x^2) dx

with limits of integration 0 and 4

volume is cubic units.

I am stuck on the last step. For some reason I tried integrating it and then plugging in the value, which in this case is just the beta, 4.

What am I doing wrong? this is supposed to be easy step..
I don't know ... you haven't shown any work.

$\displaystyle V = 2\pi \int_0^4 (x+4)(4x-x^2) \, dx$

expand the integrand and combine like terms ...

$\displaystyle V = 2\pi \int_0^4 16x - x^3 \, dx$

can you finish?

3. Thanks!

I am trying it on the other problems but got stuck on this one..

integral of 2pi x^(1-6) from 0 to 7

I got -1/4 x^4. I plugged in the 7 to get 2pi-(1/(4* 7^4)) but that's a negative volume? I took off the negative sign but still not right... thanks

4. Originally Posted by softballchick
Thanks!

I am trying it on the other problems but got stuck on this one..

integral of 2pi x^(1-6) from 0 to 7

x^(1-6) = x^(-5) ... ???

I got -1/4 x^4. I plugged in the 7 to get 2pi-(1/(4* 7^4)) but that's a negative volume? I took off the negative sign but still not right... thanks
why don't you post the entire problem as written ... ?

5. The volume of the solid obtained by rotating the region enclosed by

y=1x 6y=0 x=1 x=7

about the y-axis can be computed using the method of cylindrical shells via an integral

6. Thanks, I figured it out! I made a dumb mistake, I took the integral from 0 to 7 instead of 1 to 7.