Recent content by Quatros

I really need help on this.

(forgot to add a +3 in sign) With wolfram (we were allowed to use tech for this one) I'm getting -.0116111 ( which is the correct answer in the book). But with my parameters, -.0200198 with those parameters. y=.5x to 1, and x = 1 to 2. )

Homework Statement I'm trying to change the bounds for this integral Sin(x^2)dxdy With x going from 1 to 2y, y going from 0 to 1 (I already know the integration for sin(x^2) The Attempt at a Solution I converted 2y=x to 1/2x=y and graphed all the bounds. I went with 1,2 for my...

Ohhh, I see then i get 12cos(theta) = 1+ rcos^2(theta), then i just solve for r?

r^2(sin(theta)^2-2r^2cos(theta )^2) = r(12cos(theta)) 12cos(theta) = (sin(theta)^2 +2cos(theta)^2)r , which is where i get stuck.

I'm a bit confused, I went from y^2+2x^2 = 12x then I converted to r^2cos(theta) ^2- 2r^2sin(theta ) = 12rsin(theta) r^2(cos(theta)^2-2r^2sin(theta )) = r(12sin(theta)

All I got was to convert y = sqrt[12x-2x^2] to polar form.

Homework Statement I'm suppose to convert Sqrt[12x-2x^2] into a polar equation. Homework EquationsThe Attempt at a Solution I went from that equation to r(sin(theta)^2 + 2cos(theta)^2)= 12cos(theta), I really don't know where to go from there.

Is 8 pi correct?

Is this graph an Ice cream com This is an ice cream cone, so the ball part (theta) is going from 0 to 2pi , P/R: 0 to 2, The third parameter is giving me trouble.

yeah, I did that and still confused. Sorry about my bad english.

P 0 to sqrt(4-p^2) Theta from 0 to 2pi? the other symbol from 2 to 2?

Alright, I changed Sqrt[x^2+y^2+z^2] to p * p^2 * sin(theta) Would the z= sqrt(4-x^2+y^2) turn into sqrt(4-p^2)?

It's p for sphere our in our book, unless you aren't talking about integrand.

but isn't that cylinderical cordinates.