Surface Integrals — Question 1

Question 1 – Solution

Strategy. Treat the plane as a graph, so the same constant area factor converts every integral over $S$ into an integral over $R$.

Step 1: Surface element For $z=g(x,y)=2x+y$, $$g_x=2,\qquad g_y=1,$$ and therefore $$\boxed{dS=\sqrt{1+g_x^2+g_y^2}\,dA=\sqrt 6\,dA}.$$

Step 2: Area $$\operatorname{area}(S)=\iint_R\sqrt 6\,dA =\sqrt 6(1)(2)=\boxed{2\sqrt 6}.$$

Step 3: Weighted integral On $S$, $x+z=3x+y$. Hence $$\begin{align*} \iint_S(x+z)\,dS &=\sqrt 6\int_0^1\int_0^2(3x+y)\,dy\,dx\\ &=\sqrt 6\int_0^1(6x+2)\,dx=\boxed{5\sqrt 6}. \end{align*}$$ The average value is $$\boxed{\frac{1}{\operatorname{area}(S)}\iint_S(x+z)\,dS=\frac 52}.$$

Verification The average $5/2$ lies between the minimum $0$ and maximum $5$ of $3x+y$ on $R$.