Posted by **Ruth** on Saturday, August 12, 2006 at 4:55am.

Determine the volume of the solid abtained when the region bounded by y=sqrt x and the line Y=2 and x=0 is rotated:

(i) about the x-axis

(ii) about the line x=4.

(i) Each element of the body of revolution with thickness dx, has area

pi*(sqrt x)^2 dx = pi x dx.

Integrate that from x=0 to x = 4. (The sqrt x curve ends at x=4, y=2.) I get a volume of 8 pi.

(ii) The x=4 vertical line is at the end of the segment of the curve from x=0 to x=4. Perform an integration aloing the y axis. Each slab of thickess dy has volume

pi (4-x)^2 dy. Substitute y^2 for x and the integral becomes

(INTEGRAL sign) pi (4- y^2)^2 dy from y=0 to y=2.

## Answer this Question

## Related Questions

- Calculus - This problem set is ridiculously hard. I know how to find the volume ...
- calculus - Find the volume of the solid generated by revolving the region about ...
- Calculus - Find the volume of the solid generated by revolving the following ...
- Calculus - Find the volume of the solid generated by revolving the region ...
- calculus - Find the volume of the solid generated by revolving the region about ...
- calculus - The region bounded by the curve y=1÷(1+2x) , the line X=2 , the x-...
- Calculus - Let R be the square region bounded by y = 2, x = 2, and the x and y-...
- Calculus check - The functions f and g are given by f(x)=sqrt(x^3) and g(x)=16-...
- Calc - The region bounded by y=2.5x^2 and y=4x is to be rotated about both axes ...
- Calculus AP - Let R be the region in the first quadrant bounded by the graph y=3...