Find the area of the surface of the portion of the plane $z = 3 - 6 x - 2 y$
in the first octant.
A) $\frac { 3 \sqrt { 3 } } { 8 }$
B) $\frac { 3 \sqrt { 41 } } { 8 }$
C) $\frac { 3 \sqrt { 6 } } { 8 }$
D) $\frac { 8 \sqrt { 41 } } { 3 }$
E) $\frac { 3 } { 8 \sqrt { 41 } }$

Question

Quizplus · Accepted Answer

The surface area $A$ of a portion of the plane $z = 3 - 6x - 2y$ in the first octant can be found by integrating over the region in the $xy$-plane that projects to the first octant. The plane intersects the $z$-axis at $z = 3$, the $x$-axis at $x = \frac{1}{2}$, and the $y$-axis at $y = \frac{3}{2}$. The differential area $dA$ in the $xy$-plane is given by $\sqrt{1 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2}dxdy$, where $\frac{\partial z}{\partial x} = -6$ and $\frac{\partial z}{\partial y} = -2$. Thus, $dA = \sqrt{1 + 36 + 4}dxdy = \sqrt{41}dxdy$. Integrating over the triangle in the $xy$-plane gives the area of the surface in the first octant. The area of the triangle in the $xy$-plane is $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{8}$, so the surface area is $\frac{3}{8} \cdot \sqrt{41}$, which matches option B.

Find the Area of the Surface of the Portion of the Plane