Finding a volume bounded by a region

[asd1249jf]

Homework Statement

The region bounded by $y=e^-x^2$, y=0, x=0 and x=1 is revolved around the y-axis. Find the volume.

Homework Equations

$V = \int_{a}^{b} R^2 dy$

The Attempt at a Solution

We must first express x in terms of y. So we get

$\ln y = -x^2$

$x^2 = -\ln y$

We substitute this to the above volume equation and we get

$V = \int_{0}^{1} -\ln y$

And without even doing the integration, I know for a fact that this cannot work out since if you integrate a natural log, you get a natural log back and substituting 0 into a natural log gets you an undefined answer.

I think I did something wrong or set up the problem wrong. Any advice would be appreciated!

Oh and if you know latex, could you tell me how to express y=e^(-x^2)?

 

[Mark44]

Homework Statement

The region bounded by $y=e^-x^2$, y=0, x=0 and x=1 is revolved around the y-axis. Find the volume.

Homework Equations

$V = \int_{a}^{b} R^2 dy$

This "relevant" equation isn't relevant.

The Attempt at a Solution

We must first express x in terms of y. So we get

$\ln y = -x^2$

$x^2 = -\ln y$

We substitute this to the above volume equation and we get

$V = \int_{0}^{1} -\ln y$

And without even doing the integration, I know for a fact that this cannot work out since if you integrate a natural log, you get a natural log back and substituting 0 into a natural log gets you an undefined answer.

I think I did something wrong or set up the problem wrong.

Yes. The easiest way to do this problem is to use cylindrical shells.

The relevant equation is
$$V = \int_a^b 2\pi~radius * height~dx$$

Here, radius = x, and height = y = e^-x2

Any advice would be appreciated!

Oh and if you know latex, could you tell me how to express y=e^(-x^2)?

Put the exponent in braces:{-x^2}.
$$y = e^{-x^2}$$

 

[asd1249jf]

Just so that I understand, why is it that the cylindrical shell method must be applied here? I was taught that when finding a volume revolved around X or Y axis, the above equation works well except you just have to be careful when you plug in for X or Y.

 

[Mark44]

You could use disks in this problem, but it's more complicated. You will need two integrals to do the problem this way. For y in [0, 1/3], the disks all have a radius of 1. For y in [1/e, 1] the radius of a given disk is the x value on the curve, or +sqrt(-ln(y)).

I should add that in your relevant formula in post 1, pi is missing.