substitution on multiple integrals

Question

Let $Q$ be the region in $\mathbb{R}^2$ enclosed by the quadrilateral with vertices $(2, 4), (6, 3), (8, 4)$, and $(4, 8)$. Evaluate the double integral

$$\iint_Q \frac{5y-x}{y^2(y-2)^2}dxdy$$

I was trying to do a substitution but it did not work how can I tackle this question?

Answer 1

So it turns out that you just kind of bash your head into this integral with partial fractions and bravery. Split the quadrilateral into two triangles on the line $y=4$, and call the upper one $U$ and the lower one $L$. Then your integral becomes:

$$\iint_U \frac{5y-x}{y^2(y-2)^2}dxdy ~~+~~ \iint_L \frac{5y-x}{y^2(y-2)^2}dxdy.$$

$$=\int_{4}^{8}\!\!\int_{y/2}^{12-y} \frac{5y-x}{y^2(y-2)^2}dxdy ~~+~~ \int_{3}^{4}\!\!\int_{18-4y}^{2y} \frac{5y-x}{y^2(y-2)^2}dxdy.$$

Despite the scariness of these integrands, they are actually just linear polynomials in $x$, i.e. $a+bx$ where $a$ and $b$ are allowed to depend on $y$. So the inner integration is very easy:

$$=\int_{4}^{8} -\frac{9(7y^2-64y+64)}{8y^2(y-2)^2}dy ~~+~~ \int_{3}^{4} \frac{18(2y^2-9y+9)}{y^2(y-2)^2}dy.$$

At this step we apply partial fractions. A little bit of thought or a lot algebra will convince you that the "linear-denominator" terms will vanish, and indeed

$$=\frac{9}{8}\int_{4}^{8} -\frac{16}{y^2}+\frac{9}{(y-2)^2}dy ~~+~~ 18\int_{3}^{4}\frac{9/4}{y^2}-\frac{1/4}{(y-2)^2}dy.$$

And again, we now just have four integrals, each of which is computed with the power rule.

$$=\frac{9}{8} ~~+~~ \frac{9}{8}$$

Therefore, the integral is $\frac{9}{4}$.

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Having seen the solution long-hand, it's interesting to me that (1) the answer is nice, (2) because each of the two halves have nice answers, instead of there being some gross terms which cancel between the two, and (3) each half has the same nice answer.

Therefore, I wouldn't be surprised if there is a nice change of variables $(x,y)\mapsto(u,v)$ which both

• simplifies the integrands somewhat and also
• makes the two triangles congruent in the $(u,v)$-plane.

But, as a general principle, there are lots of changes of variables out there, and I don't think it's wise to sit around hoping that you're going to come up with some magic one for any particular integral. Better just to put in the elbow grease :)