The parametric surface has the following parametric equation:
$$f(r,\theta)=(x,y,z)=(2r\cos\theta,5r\sin\theta,r).$$ Then we have
$$f_r(2,\pi/4)=(2\cos(\pi/4),5\sin(\pi/4),1)=(\sqrt{2},5\sqrt{2}/2,1),$$
$$f_\theta(2,\pi/4)=(-2\cdot 2\cdot\sin(\pi/4),5\cdot 2\cdot\cos(\pi/4),0)=(-2\sqrt{2},5\sqrt{2},0).$$
Therefore, the unit normal $n$ of the tangent plane at $(r,\theta)=(2,\pi/4)$ is equal to
$$n=\frac{f_r(2,\pi/4)\times f_\theta(2,\pi/4)}{\|f_r(2,\pi/4)\times f_\theta(2,\pi/4)\|}=\frac{1}{\sqrt{458}}(-5\sqrt{2},-2\sqrt{2},20).$$
Since the tangent plane at $(r,\theta)=(2,\pi/4)$ passes through the point $f(2,\pi/4)=(2\sqrt{2},5\sqrt{2},2)$, the equation of the tangent plane is given by
$n\cdot(x-2\sqrt{2},y-5\sqrt{2},z-2)=0$, or equivalently,
$$(-5\sqrt{2},-2\sqrt{2},20)\cdot(x-2\sqrt{2},y-5\sqrt{2},z-2)=0,$$
that is
$$5\sqrt{2}x+2\sqrt{2}y-20z=0.$$
$z=1-x-y$ so set $f(x,y)=1-x-y$ $$S=\int\int_D\sqrt{f_x^2+f_y^2+1}dxdy$$
$$S=\int\int_D\sqrt{3}dxdy$$
$$S=\sqrt{3}\int\int_Ddxdy$$
$$S=\sqrt{3}(4\pi) $$
$$S=4\sqrt{3}\pi$$
Note: $D$ is the region enclosed by the circle $x^2+y^2=4$ in $x,y$ plane.
Best Answer
It is the circle $x^2+z^2=2a^2$ in the plane $y=a$
$\left(a\sqrt{2} \cos t,a,a\sqrt{2} \sin t\right);\;t\in [0,2\pi]$
It's area is $2a^2\pi$