Volume of a Solid

[cocomo]

I need to find the volumes of a solid in the region bounded by y=1, x=5, and y=e^x, when rotated around A) x=5, B) y=e^8, C) y=1. I need help writing the integral expressions. I am familiar with the formula for finding the volume of a solid, but have never done this with an exponential function.

 

[galactus]

To revolve about x=5, we can use shells or washers.

Washers:

In this case, we integrate w.r.t y. This changes the integration limits from 0 to 5 to $\displaystyle e^{0}=1 \;\ \text{and} \;\ e^{5}$

$\displaystyle {\pi}\int_{1}^{e^{5}}(ln(y)-5)^{2}dy$

Shells:

When using shells, the cross-sections are parallel to the axis about which we are revolving. Since we are revolving about x=5, the cross-sections are parallel to x=5. Picture them stacked up along the x-axis from 0 to 5. That is why we integrate w.r.t x in this case.

$\displaystyle 2{\pi}\int_{0}^{5}(5-x)(e^{x}-1)dx$

You will get the same solution with both set ups.

Can you get the others by using this as a template?.

 

[BigGlenntheHeavy]

Here are the three formulas, see if you can figure them out.


$\displaystyle A) \ = \ \ \pi \int_{1}^{e^{5}}(ln|y|-5)^{2}dy, \ disc$
$\displaystyle B) \ = \ \ \pi \int_{0}^{5}[(1-e^{8})^{2}-(e^{x}-e^{8})^{2}]dx, \ washer$
$\displaystyle C) \ = \ \ \pi\int_{0}^{5}(e^{x}-1)^{2}dx, \ disc$

 

[galactus]

Here is B using shells:

$\displaystyle 2{\pi}\int_{1}^{e^{5}}(y-e^{8})(ln(y)-5)dy$

C is basically the same set up, except with 1 instead of e^8.

 

[cocomo]

Thank you very much for your help. Can you explain why for the first one it is lny-5 and not 5-lny? I thought 5 would be the outside radius and then subtract the inside radius, lny...?