Find the surface area of the portion of the plane z = -2x + 4y + 1 lying above the region 0 <= x <= 3, 0 <= y <= 5. Give an exact answer.

Step-by-step solution

1. 1

   Recall the surface-area formula. For a graph $z=f(x,y)$ over a region $R$, $S=\iint_R\sqrt{1+f_x^{\,2}+f_y^{\,2}}\,dA.$

2. 2

   Differentiate the plane. With $f(x,y)=-2x+4y+1$, $f_x=-2$ and $f_y=4$ everywhere. Because $f$ is linear, these are constants — that is what makes the integral trivial.

3. 3

   Evaluate the integrand. $\sqrt{1+(-2)^2+4^2}=\sqrt{1+4+16}=\sqrt{21}.$ Note the sign of $f_x$ is irrelevant once it is squared.

4. 4

   Pull the constant out. $S=\iint_R\sqrt{21}\,dA=\sqrt{21}\iint_R dA=\sqrt{21}\cdot\operatorname{area}(R).$

5. 5

   Compute the base area and finish. The rectangle has area $3\times 5=15$, so $S=15\sqrt{21}\approx 68.7386.$ Geometrically, $\sqrt{21}$ is the constant factor by which a tilted plane stretches horizontal area.