What's the point of intersection for y = x^1/2 and y = e^-3x in the interval (0,1)?

Question

I was asking to find the rigion bound by y = x^1/2 , y =  e^-3x and x=1.  And the volume revolving it around the x-axis.

Answer 1

The point of intersection is given by x^1/2=e^-3x. It turns out that x=0.23873 approx.

To find the volume of revolution, you need a picture of the graphs so the you can see what you are doing.
 

Integral of ydx between x=0 and 1 I'll write as S[0,1](ydx). This would give you the area enclosed.

Volume of rev=(pi)S[0,1](y^2dx) the sum of the volumes of thin discs about the x axis.

Up to the point of intersection, the volume is governed by y=x^1/2, and after that it's the other function.

(pi)S[0,0.23873](y^2dx)=(pi)S[0,0.23873](xdx)=(pi)x^2/2[0,0.23873]=0.08952 (first part of integration)

(pi)S[0.23873,1]((e^-6x)dx)=(pi)(-e^-6x)/6[0.23873,1]=(0.23873/6-e^-6/6)(pi)=0.12370 (second part of integration)

Note that e^-6x=(e^-3x)^2=x=0.23873 at the intersection.

Total volume=0.08952+0.12370=0.2132 approx.