Thread: 2 volumes question

1. find the volume when the region bounded by the curves $\displaystyle y= x^2$ and $\displaystyle y= \frac{1}{4}x^2 + 1 $ in the 1st quadrant is rotated about the line x = 2/root 3

2. determine the volume of rotating the region bounded by $\displaystyle y = e^{-x^2} $ and the coordinate axes about the y axis.

I have no idea how to do Q1 but for Q2 I got pi cubic units. Is that correct?

Thanks

in the first question i think it would help you a lot if you sketched these curves.
After youve done that, find the points where the two curves intersect.
$\displaystyle x^2=\frac{1}{4} x^2+1$

now when you have the two points you notice they are on the line you will be revolving your curves around.
I found it useful to change the position of the x axis to be where that line is. This way the problem comes down to revolving the curves around the x-axis.

The volume of revolution about the x axis is easy enough to do, all you need to remember here is that you revolve two curves and you want their common volume.
Thats why you need to subtract one of the volumes from the other, this will give you the value you're looking for.
Oh and dont forget about halfing the result, it is symmetrical about the y axis so what you get is in 1st and 2nd quadrant.


solving the second question comes down to noticing that if

$\displaystyle y= e^{-x^2}$
then
$\displaystyle x^2=-ln(y)$

now you have a ready expression to substitute to the formula, the limits of integration here you get by setting x=0 and y=0 and seeing what happens to the function.
i think the limits should be from 0 to 1 and that gives you the answer V=pi

Thanks for contributing. I still didn't get Q1 though

why dont you show your working and i can point you in the right direction?