SOLUTION: please help me solve this problem. Find the volume generated by revolving the area in the first quadrant under the curve y=x^2-x^3 about the y-axis? Thank You!!!

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: please help me solve this problem. Find the volume generated by revolving the area in the first quadrant under the curve y=x^2-x^3 about the y-axis? Thank You!!!       Log On


   



Question 717838: please help me solve this problem.
Find the volume generated by revolving the area in the first quadrant under the curve y=x^2-x^3 about the y-axis?
Thank You!!!

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
That area in the first quadrant looks like this:



and when it's rotated about the y-axis its cross section will look
like this:



We will use the cylindrical shell method:



The volume of a cylindrical shell is:
 
(its circumference) times (its height) times (its thickness).

Its circumference is 2πr and its thickness is dx.  Its radius, r, is the x
value of the point (x,y) which is just x. Its height, h, is just the y
value of the point(x,y) which is just y, (which we will replace by x²-x³):  

V = int%282pi%2Ar%2Ah%2Adx%2C%22%22%2C0%2C1%29%22%22=%22%222pi%2Aint%28r%2Ah%2Adx%2C%22%22%2C0%2C1%29%22%22=%22%222pi%2Aint%28x%2Ay%2Adx%2C%22%22%2C0%2C1%29%22%22=%22%222pi%2Aint%28x%28x%5E2-x%5E3%29%2Adx%2C%22%22%2C0%2C1%29%22%22=%22%222pi%2Aint%28%28x%5E3-x%5E4%29%2Adx%2C%22%22%2C0%2C1%29%22%22=%22%22

matrix%283%2C2%2C%0D%0A%22%7C%22%2C1%2C%0D%0A%22%7C%22%2C%22%22%2C%0D%0A%22%7C%22%2C0%29%22%22=%22%22%282pi%29%28x%5E4%2F4-x%5E5%2F5%29matrix%283%2C2%2C%0D%0A%22%7C%22%2C1%2C%0D%0A%22%7C%22%2C%22%22%2C%0D%0A%22%7C%22%2C0%29%22%22=%22%22%282pi%29%281%5E4%2F4-1%5E5%2F5%29%22%22-%22%22%282pi%29%280%5E4%2F4-0%5E5%2F5%29%22%22=%22%22%282pi%29%281%2F4-1%2F5%29%22%22=%22%22
%282pi%29%285%2F20-4%2F20%29%22%22=%22%22%282pi%29%281%2F20%29%22%22=%22%22pi%2F10   

Edwin