as I see it, the upper and lower boundries cross at 5,25, os the integral will be from x=0 t0 5
so the dArea will be [(50-x^2)-x^2]dx
and you will rotate that about the y axis, so the volume will be INT 2PI*xdA or
int 2PI x(50-2x^2)dx
so integrate that. I will be happy to check it.
Find the volume of the solid generated by revolving the region bounded by the given curves and line about the y-axis.
y=50-x^2
y=x^2
x=0
3 answers
Like all multiple integral problems, start with drawing a sketch of the bounding curves.
I have done that for you for this time, see:
http://img687.imageshack.us/img687/811/1342739949.png
If curves intersect, find the intersection points. In this case, it is at (5,25).
Then decide how you want to integrate, namely the order of integrating along x first, followed by y, or vice versa.
Using the ring method, and set up the double integral:
Volume
=∫∫2πx dy dx
y goes from x^2 to 50-x^2 and
x goes from 0 to 5 (intersection point).
I have done that for you for this time, see:
http://img687.imageshack.us/img687/811/1342739949.png
If curves intersect, find the intersection points. In this case, it is at (5,25).
Then decide how you want to integrate, namely the order of integrating along x first, followed by y, or vice versa.
Using the ring method, and set up the double integral:
Volume
=∫∫2πx dy dx
y goes from x^2 to 50-x^2 and
x goes from 0 to 5 (intersection point).
2Pi(500/3)
0 answers