Write a double integral for surface area of cone $z=1-\sqrt{x^2+y^2}$ above the plane $x+2z+1=0$.
First I started with finding projection of these two in $xOy$ plane: $-\frac{1}{2} - \frac{x}{2}=1-\sqrt{x^2+y^2}$ ... which is ellipse $$\frac{(x-1)^2}{2^2} + \frac{y^2}{(\sqrt{3})^2}=1.$$ Then using formula $$P=\iint_D \sqrt{1+p^2+q^2}dxdy;$$ I have $z(x,y)=1-\sqrt{x^2+y^2}$ I can easily find partials $p$ and $q$ and I get $\sqrt{2}$ for the $P=\int\int_D \sqrt{2}dxdy$.
Now the area over which the integration is done $D$:
Since it is ellipse I put elliptical coordinates to be: $x=1+2rcost$ and $y=\sqrt{3}rsint$, where $0\leq r \leq 1$ and $0\leq t \leq 2\pi$ and Jacobian is $2\sqrt{3}r$.
Final, the integral would look like this: $$P=\int_{0}^{2\pi}\int_{0}^{1} \sqrt{2}2\sqrt{3}rdrdt.$$
Is this correct?