Let r be the region bounded by y=x and y=x^2.find the area of r. find the volume of the solid that results when r is revolved about the x-axis(using the shell method).

2pi/15

Step-by-step explanation:

The intersection of y=x and y=x^2 is (0,0) and (1,1)

Washer method

Prefered this method for this one since slices are perpendicular to axis of rotation

Integrate(Big circle - Small Circle)dx

Integrate(pi*(x)^2-pi*(x^2)^2)dx on x=0..1

Integrate(pi*x^2-pi*x^4)dx on x=0..1 is 2pi/15

Shells method

So I have to solve for x since I will be integrating with respect to y

The height of the the shell will be sqrt(y)-y

The radius will be y since that is the distance between axis of rotation in a point within the shell

Integrate(2*pi*r*h ) dy

Integrate(2*pi*(sqrt(y)-y)*y) dy on y=0..1 is 2pi/15

4x and 6y

im not sure though

the two may have that, even a breakdown is probably offered