Tuesday

September 30, 2014

September 30, 2014

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

(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.

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