Iterated Integrals — Question 7

Question 7 – Solution

Strategy. Integrate the three terms separately, divide by plate area, and then solve the resulting level equation at convenient choices of $y$.

Step 1: Total and average The plate area is $A=3\cdot 2=6$. Its total temperature integral is $$\begin{align*} \int_0^3\int_{-1}^{1}(18+2x+y^2)\,dy\,dx &=18(6)+\left(\int_0^3 2x\,dx\right)(2) +3\int_{-1}^{1}y^2\,dy\\ &=108+18+2=128. \end{align*}$$ Hence $$\boxed{T_{\mathrm{avg}}=\frac{128}{6}=\frac{64}{3}}.$$

Step 2: Range check The minimum is $T(0,0)=18$. The maximum $25$ occurs at $(3,1)$ and $(3,-1)$. Since $$18<\frac{64}{3}<25,$$ the average lies strictly inside the temperature range, as required.

Step 3: Points at the average Solve $$18+2x+y^2=\frac{64}{3}.$$ Taking $y=0$ gives $x=5/3$, while taking $y=1$ gives $x=7/6$. Thus two examples are $$\boxed{\left(\frac 53,0\right)\quad\text{and}\quad\left(\frac 76,1\right)}.$$ Both lie on the plate, and substitution gives $64/3$ at each.