Answer
$6\sqrt 6-2 \sqrt 2$
Work Step by Step
Here, we have $x^2-2y-2z=0$
As we know that $x=r \cos \theta, y=r \sin \theta $
The surface area is given by:
$S=\iint_{R} \dfrac{|\nabla f|}{|\nabla f \cdot p|} dA$
This implies that
$S=\int_{0}^{2} \int_{0}^{3x}\sqrt {x^2+2} dx dy$
Thus, $S=\int_{0}^{2} 3x \sqrt {x^2+2} dx=6\sqrt 6-2 \sqrt 2$
