Did i do this correctly

1. Mar 27, 2005

Let R be the region in the first quadrant bounded by the y-axis,y=2 and the graph of $$e^\frac{x}{2}$$. Create a solid of revolution by revolving R about the y-axis. Use the shell method to find the volume. This is the integral i've come up with, is it correct? And of coures i'll use inegration by parts to calculate the integral.

$$2\pi\int_{0}^{\ln4}(x)(e^\frac{x}{2})dx$$

and using integration by parts yields.

$$2\pi(2xe^\frac{x}{2}-4e^\frac{x}{2})\right]]_{0}^{\ln4}$$

Last edited: Mar 27, 2005
2. Mar 27, 2005

TimNguyen

Looks correct.

3. Mar 27, 2005

thx timnguyen. can someone else please check this solution for me too?

4. Mar 27, 2005

Zurtex

Yep that's exactly right.

5. Mar 27, 2005

sorry guys this is not correct!!!!!

6. Mar 28, 2005

the integration of the integral is correct, but for this particular problem the integral is wrong.

7. Mar 28, 2005

whozum

I don't see anything wrong with that integral. Make sure you evaluated it right.

I get 9.7086

8. Mar 28, 2005

the calculation of the integral is correct BUT, it is not the correct integral for the problem. from the information given the integral should be

$$2\pi\int_{0}^{\ln4}(2x-xe^\frac{x}{2})dx$$

9. Mar 28, 2005

whozum

so $$f(x) = 2-e^{x/2}$$? I don't see why.

For cylindrical shells:

V = $$2\pi\int{xf(x)}{dx}$$

f(x) is given in the problem as $$e^{x/2}$$

$$2-e^{x/2}$$ is the original function reflected over y=1.

10. Mar 28, 2005

HallsofIvy

Staff Emeritus
Because using the shell method, you use the fact that the surface area of a cylinder of radius r and height h is $$2\pi r^2 h$$. Here r= x and height of the "cylinder" is from $$e^{\frac{1}{2}x}$$ to 2- the length is $$2- e^{\frac{1}{2}x}$$.

11. Mar 28, 2005

dextercioby

Halls,i hope u mean

$$S_{lat.right circular cyl.}=2\pi r h$$

Daniel.

12. Mar 28, 2005

Zurtex

Oh well, I've never used the shell method, I only just looked it up on MathWorld. Guess that'll help me if I ever come across it