# volumes of revolution

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Thread starter 15 years ago
#1
find the region enclosed when the following curves and the axis is rotated through 360 degrees about the x axis find the volume of the solid generated

a) y=(x+1)(x-3)

b) y=1-x^2

keep gettin the wrong answer!! if someone could do them would b great!!! thanks
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15 years ago
#2
no problem.... volume = pi*integral(y^2)

y=(x+1)(x-3)
so... [(x+1)(X-3)]^2=X^4-4X^3-2X^2+12X+9
so... integral = 1/5X^5-X^4-2/3X^2+6X+9X between -1 and 3... 34.133
then times integral value by pi... so volume = 107.232 approx

y=1-x^2
(1-x^2)^2=X^4-2X^2+1
so integral = 1/5X^5-2/3X^3+X.. limits -1, 1
so value = 1.066666 times pi.... volume = 3.351 approx

if anything is wrong.. I apologise... my P2 is a bit rusty...
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