Question

Question: What is the volume of the revolution bounded by the curves of y=4-x^2 , y=x, and x=0 and is revolved about the vertical axis.

First, I had found the points of intersection to get the limits and I got -2.5616 and 1.5616. And then I plug it in the shell method formula and I got the lower limit of -2.5616 to upper limit of 1.5616 of the integral of x((4-x^2)-x)dx times all of that by 2pi. I was wondering if I did this all wrong ?

Answer 1

I guess the decimal values are ok, but I tend to prefer the exact values of (-1+√17)/2 and (-1-√17)/2. They are a bit cumbersome to work with by hand, but with the wealth of online web sites to do the calculations, that should be no obstacle.

Your limits of integration are wrong, since x=0 is one boundary. Including area left of the y-axis would involve overlap during the revolution.

So, using shells of thickness dx, the volume

v = ∫[0,(-1+√17)/2] 2πrh dx
where r=x and h=(4-x^2)-x
v = ∫[0,(-1+√17)/2] 2πx(4-x^2-x) dx

Your integrand looks good.

Answer 2

Thanks a bunch Steve!