Is This the Correct Integral for Finding Volume Using the Shell Method?

[RadiationX]

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}

 

[TimNguyen]

Looks correct.

 

[RadiationX]

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

 

[Zurtex]

Yep that's exactly right.

 

[RadiationX]

sorry guys this is not correct!

 

[RadiationX]

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

 

[whozum]

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

I get 9.7086

 

[RadiationX]

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

 

[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.

 

[HallsofIvy]

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}.

 

[dextercioby]

Halls,i hope u mean

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

Daniel.

 

[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