find surface area using double integral of: $\sqrt{(x^2+y^2)}+z=1$ and plane $x+2z+1=0$ So I have to find surface area using double integral of: $\sqrt{(x^2+y^2)}+z=1$ above the plane $x+2z+1=0$. I drew the picture and I see that an ellipse is the intersection between the cone and plane.
Now, $z(x,y)=1-\sqrt{(x^2+y^2)}$ and from $z$ I will find partial derivatives by $x$ and $y$ and put them in a standard formula. Now what is giving me trouble is when I have to project into the x-y plane. I can't seem to understand over which area should I integrate.. Is it the ellipse?
------------------------EDIT: based on Math Lover comment------------------------
So I expressed $z$'s from cone and plane equation and set them equal in order to get the ellipse equation. At the end I got $\frac{(\sqrt{3}x-9)^2}{18}+\frac{y^2}{\frac{9}{2}}=1$. Then I performed the change of variables $x=\frac{9+\sqrt{18}u}{\sqrt{3}}$, $y=\frac{\sqrt{9}}{\sqrt{2}}v$ in order to get unit circle in $u-v $ coordinate system. Also I have found the Jacobian $\frac{\partial(x,y)}{\partial(u,v)}$. Finally at the end I used polar coordinates and that's it. Is this the right way to do so?
 A: Something went wrong in one of your steps. I get a different equation for ellipse.
Parametrization of the cone, $ \ c(t) = (r \cos t, r \sin t, 1 - r)$.
So you can easily find $|c'_{t} \times c'_{r}| = r \sqrt2$
So surface area of the cone $S = \sqrt2 \iint_D r \ dr \ dt$
Where $D$ is the projection of the intersection of the cone and the plane in $XY$ plane. So our next step is to find the area of the intersection.
Intersection of cone and the plane is given by,
$1 + \frac{1+x}{2} = \sqrt{x^2 + y^2} \implies 3x^2 - 6x + 4y^2 = 9$
So equation of ellipse is $ \ \frac{(x-1)^2}{4} + \frac{y^2}{3} = 1$
We know the area of the ellipse with semi-major and semi-minor axes lengths of $a$ and $b$ is $\pi ab = 2 \sqrt3 \ \pi$
But if you need to solve this by integral then,
Using substitution, $x = 1 + 2 \ r \cos t, y = \sqrt3 \ r \sin t$, our new domain is a circle of unit radius.
$|J| = 2 \sqrt3 \ r$
So the area of ellipse $\displaystyle \iint_D dx \ dy = \int_0^{2\pi}\int_0^1 2 \sqrt3 \ r \ dr \ dt = 2 \pi \sqrt3$
So surface area of the cone $S = 2 \pi \sqrt6$
