How do you find the volume bounded by y^2=x^3y2=x3 and y=x^2y=x2 revolved about the y-axis?

1 Answer
Jun 10, 2018

pi/20π20 units ^3

Explanation:

We need to find where these two curves intersect to find the bounds of integration.

y^2=x^3 and y=x^2y2=x3andy=x2, squaring the second expression, y^2=x^4y2=x4 Solving for y^2y2,.......... [x^4=x^3x4=x3] i.e, x^3[x-1]=0x3[x1]=0.

So x=1, x=0x=1,x=0 are the points of intersection.

From the graphs of these expressions in can be seen that y=sqrt[x^3]y=x3 has a greater area than y= x^2y=x2 so we must find the area under y =x^2y=x2 and subtract it from the area under y=sqrt[x^3]y=x3 and then revolve this area about the xx axis between the bounds x=1, x= 0x=1,x=0

Volume of revolution is given by piint_a^by^2dxπbay2dx So the volume of revolution = [piint_0^1x^3dx[π10x3dx - piint_0^1x^4dx]π10x4dx].= pi[x^4/4- x^5/5]π[x44x55] [ after integration by the general power rule] and evaluated for x=1, x=0x=1,x=0 will result in pi[ 1/4-1/5]π[1415] = pi/20π20.