- #1

- 30

- 0

y=x^2

y=2x

about the x-axis.

So i used intergral 0 to 2 (2x-x^2)^2 dx

this is wat i got....

pi*(((2^5)/5)-(4(2^4)/4)+(4(2^3)/3))

its wrong so wat did i do wrong?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter zcabral
- Start date

- #1

- 30

- 0

y=x^2

y=2x

about the x-axis.

So i used intergral 0 to 2 (2x-x^2)^2 dx

this is wat i got....

pi*(((2^5)/5)-(4(2^4)/4)+(4(2^3)/3))

its wrong so wat did i do wrong?

- #2

- 1,341

- 3

That region will make a washer. Your interval [0,2] looks right.

However it might be beneficial to find a small piece of volume of the actual shape.

V=A*h so an infinitely small piece of volume would be dV = A*dh, or more approately in our case:

dV=A*dx

The area is pi*r^2 since it forms a washer, so [tex]A=\pi (r_1^2-r_2^2)[/tex]

Where r1 and r2 are the distances from the x-axis where r1>r2

The Area portion is what you did wrong, do you see where to go from there?

Next time you should post questions like this (non-conceptual) questions in the homework help.

However it might be beneficial to find a small piece of volume of the actual shape.

V=A*h so an infinitely small piece of volume would be dV = A*dh, or more approately in our case:

dV=A*dx

The area is pi*r^2 since it forms a washer, so [tex]A=\pi (r_1^2-r_2^2)[/tex]

Where r1 and r2 are the distances from the x-axis where r1>r2

The Area portion is what you did wrong, do you see where to go from there?

Next time you should post questions like this (non-conceptual) questions in the homework help.

Last edited:

Share: