# Math Help - Let R be the region ... find volume when R is revolved around x-axis

1. ## Let R be the region ... find volume when R is revolved around x-axis

let R be the region in the first quadrant between the graphs of $y=e^{-x}$, $y= sinx$ and the y-axis. The volume of the solid that results when R is revolved about the x-axis is

(A) -0.888
(B) -0.869
(C) 0.277
(D) 0.869
(E) 0.888

first thing i did was graph it on calculator then find where the graphs intersect but i forgot what to do from there

2. First of all, volume can't be negative so A and B are out.

3. Do you know the equation of finding the volume?

4. Originally Posted by dwsmith
Do you know the equation of finding the volume?
$\pi \int_a^b (OR)^2 - (IR)^2 dx$

where OR= the outer radius and IR= the inner radius

here's my attempt at the problem

a≈0.588532
b≈ 3.096364

$\pi \int_a^b (sinx)^2 - (e^{-x})^2 dx$

using calculator to find the definite integral i get

$\pi \int_a^b (sinx)^2 - (e^{-x})^2 dx$ ≈1.354
problem is that is not one of the options which means i did something wrong.

5. Just to verify the answer is E right?

6. Originally Posted by dwsmith
Just to verify the answer is E right?
yes the answer is E but i don't know how to get the answer

7. I think you are misinterpreting the question. The region you are looking for starts at y=0 to the intersection.

So your bounds of integration on the x axis are 0 to .588.

I think you may have done the region further to right.

8. Originally Posted by dwsmith
I think you are misinterpreting the question. The region you are looking for starts at y=0 to the intersection.

So your bounds of integration on the x axis are 0 to .588.

I think you may have done the region further to right.
thanks