Expected value of a circuit given $f_{X, Y}(x, y) = k(x + y)$

Question

HW problem, not sure where I'm going wrong on this.

Find $E(R)$ for a two-resistor circuit similar to the one described in Example 3.9.2, where $f_{X, Y}(x,y) = k(x + y)$, | $10 ≤ x ≤ 20$, $10 ≤ y ≤ 20$

Example 3.9.2 shows resistors set up in parallel, and gives the formula $\frac{1}{R} = \frac{1}{x} + \frac{1}{y}$.

So I did that formula and came up with $$R = \frac{xy}{x + y}$$

Then plugging in

$$k\int_{10}^{20}\int_{10}^{20}{\left(\frac{xy}{x + y}\right)\left(x + y\right)}dx dy$$

Which cancels to

$$k\int_{10}^{20}\int_{10}^{20}{(xy)}dx dy$$

Which gives me the answer of $45000k$. Where am I going wrong? The answer should be 7.5 according to the answers in the back. Nothing in the problem indicates that $k = \frac{1}{6000}$ that I see, nor anything in the referenced example.

Thanks.

Answer 1

HINT you have to find $k$ such that $$k\int_{10}^{20}\int_{10}^{20}{\left(x + y\right)}dx dy=1$$