use the method of disks to find the volume of the solid?

Question

enclosed by y=x and y=x^2 about the line x= -1

Answer 1

It's late, I'm sleepy, but I'll try it. The enclosed area goes from (0,0) to (1,1). You want to use disks revolving around x = -1, so you have to use dy and integrate on y. That means you need x = f(y), and so your bounds are x=y and x = sqrt y.

For the parabola, the radius is 1 + sqrt y, and the outer area is pi (1 + sqrt y)^2 = pi (1 + 2 sqrt y + y)

For the line, the radius is 1+y, and the inner area is pi (1+y)^2 = pi (1 + 2y + y^2)

To get the area of your disk, you've got to subtract the inner area from the outer one, getting

pi (1 + 2 sqrt y + y) - pi (1 + 2y + y^2)

= pi (2 sqrt y - y - y^2) = pi [ 2y^(1/2) - y - y^2]

You have to integrate that from 0 to 1. The integral is

pi [(4/3)y^(3/2) - (1/2)y^2 - (1/3)y^3]

With the lower limit, you get zero, and with the upper limit, you get

pi (4/3 - 1/2 - 1/3) = pi/2 (Answer)

Answer 2

Draw a picture of your curves, now transfer the graphs to the right by 1, so that rotation be around the line x=0 that is around Y-axis. The volume will not suffer of this transformation!
Thus y=x becomes y=x-1 and y=x^2 becomes y=(x-1)^2, and our integral I=I1+I2, where I1 is volume above X-axis, I2 is volume below X-axis.
I1) Elementary disk dv = pi*x^2 * dy = (pi*x^2) * (2*(x-1)*dx) = 2pi*x^2*(x-1)*dx; thus
I1 = integral {for x=0 until 1} of 2pi*(x^3-x^2)*dx = 2pi*(x^4/4 - x^3/3) = -pi/6; sign minus for volume does not matter, so I1=pi/6

I2) elementary disk dv = pi*x^2 * dy = (pi*x^2) * dx, thus I2 = (pi/3)*x^3 = pi/3
The whole I=pi/2

Answer 3

i hate disk method so much. I did the work and i got Pi/2 but i could be wrong.

Answer 4

dV = 2πrdrdh
r = x + 1, dr = dx
..... .... x y^1/2
V = 2π∫∫(x + 1)dydx
..... x^2 y
..... .... 1..... ..... ..... . y^1/2
V = 2π∫((1/2)x^2 + x)|dy
..... ... 0..... ..... ..... .. y
..... .... 1
V = 2π∫(-(1/2)y + y^1/2 - (1/2)y^2)dy
..... .... 0
..... ..... ..... ..... ..... ..... ..... ...... ..... ..... 1
V = 2π[(-(1/4)y^2 + (2/3)y^3/2 - (1/6)y^3)]
..... ...... ..... ..... ..... ..... ..... ..... ..... ..... 0
V = 2π[-(1/4) + (2/3) - (1/6)]
V = 2π[-(3/12) + (8/12) - (2/12)]
V = 2π(3/12)
V = π/2