Vector Calculus (area II)

Calculate the area of the cone $\displaystyle \ z = \sqrt {x^2 + y^2}$, that is inside cylinder $\displaystyle x^2 + y^2 \leq 2x$, outside the cylinder $\displaystyle x^2 + y^2 \leq 1$ and above the plane xy.

A.: $\displaystyle \frac {\sqrt{2} (2\pi + 3 \sqrt{3})} {6}$

Re: Vector Calculus (area II)

Are you sure it's not asking you to calculate the VOLUME, not the area?

Re: Vector Calculus (area II)

Yes, I'm sure. The formula is: $\displaystyle A = \iint \limits_S \left ||{{\vec{r_u}.\vec{r_v}}} \right || dA $

Re: Vector Calculus (area II)

Yes, I'm sure. The formula is: $\displaystyle A = \iint \limits_S \left ||{{\vec{r_u}.\vec{r_v}}} \right || dA $.

But I can't figure out the limits of integration.

--> I'm sorry for the repeated post (how can I delete the first one?)