# find the volume of the solid generated by revolving the region bounded by the curve y=e^(x^2) and the lines y=0 and x=1 about the y axis.

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### 1 Answer

I'm guessing you also want the region bounded by `x=0` ` `

We can do this using the "shells" method:

We want to cut our shape into a whole bunch of concentric cylinders. Each cylinder has a radius, a height, and a thickness. Think of the red line on the graph as representing one of the cylinders. (if you were to rotate it around the axis, you would get a thin cylinder)

Say that red line is "x: away from the y-axis. Then the thin cylinder it makes when rotated has a radius of x. The height of the cylinder is what you get when you plug x into the equation for y, so it is `e^(x^2)`

Thus, the volume of the cylinder made by rotating the red line about the axis is: `2 pi x e^(x^2) Delta x`

where `Delta x` is the thickness of the red line.

We want to add up all the cylinders from `x=0` to `x=1` and let the thickness go to 0. This is an integral:

`int _0^1 2pi x e^(x^2) dx `

We use a u-substitution:

`pi int 2x e^x^2 dx`

Let `u=x^2` , so `du = 2x dx` and the integral becomes:

`pi int e^u du = pi e^u = pi e^(x^2)`

Thus:

`pi int_0^1 2x e^(x^2)dx = [pi e^(x^2)]_0^1`

`=pi(e^1-e^0) = pi (e-1)`