Find the average value of the function: z(x,y) = 2x+y
Find the average value of the function: $$ z(x,y) = 2x+y $$ over the region bounded by the lines $x=0$, $y=0$, and $x+y=3$.
Explanation of Method
As in the previous problem, compute $\iint_D z\,dA$ directly, divide by the triangle's area, and cross-check against the average of the vertex values.
- Find the vertices and area of the triangle.
- Integrate $z$ over the triangle.
- Divide to get the average, and verify against the vertex-average shortcut.
Step 1 · Identify the region
The triangle has vertices $(0,0)$, $(3,0)$, $(0,3)$, with area $\operatorname{Area}(D)=\frac12\cdot3\cdot3=\frac92$.
Step 2 · Set up the integral
$$ \iint_D z\,dx\,dy = \int_0^3\int_0^{3-x}(2x+y)\,dy\,dx. $$
Step 3 · Integrate with respect to $y$
$$ \int_0^{3-x}(2x+y)\,dy = \left[2xy+\frac{y^2}{2}\right]_0^{3-x} = 2x(3-x)+\frac{(3-x)^2}{2} = 3x-\frac{3x^2}{2}+\frac92. $$
Step 4 · Integrate with respect to $x$
$$ \int_0^3\left(3x-\frac{3x^2}{2}+\frac92\right)dx = \left[\frac{3x^2}{2}-\frac{x^3}{2}+\frac{9x}{2}\right]_0^3 = \frac{27}{2}-\frac{27}{2}+\frac{27}{2} = \frac{27}{2}. $$
Step 5 · Compute the average value
$$ \bar z = \frac{1}{9/2}\cdot\frac{27}{2} = \frac{27}{2}\cdot\frac29 = 3. $$
Additional Notes
- Consistency check: $z(0,0)=0$, $z(3,0)=6$, $z(0,3)=3$, and $\frac{0+6+3}{3}=3$, matching.
Result
The solution is: