# Thread: Shape revolved around the x-axis?

1. ## Shape revolved around the x-axis?

I have this question: I found that the volume would be around 10, and I believe the last equation fits that, but my friend (in the same class) told me that it can't be around 10. Am I right or is he? If he is what did I do wrong? That's what I had when I integrated e^x-x from 0 to 1.

2. ## Re: Shape revolved around the x-axis?

Use the disk method. The radius of each circle is the distance from the x-axis to the curve y=e^x. So, the area is:

$\displaystyle \pi\int_0^1(e^x)^2dx = \pi \int_0^1 e^{2x}dx = \dfrac{\pi}{2}\left(e^2-1\right)$

So, your answer is correct. You should understand the integration technique rather than guessing.

3. ## Re: Shape revolved around the x-axis?

Originally Posted by SlipEternal
Use the disk method. The radius of each circle is the distance from the x-axis to the curve y=e^x. So, the area is:

$\displaystyle \pi\int_0^1(e^x)^2dx = \pi \int_0^1 e^{2x}dx = \dfrac{\pi}{2}\left(e^2-1\right)$

So, your answer is correct. You should understand the integration technique rather than guessing.
It wasn't a complete guess, when I integrated I had about 10, and then I checked every option. Thank you for helping!