Finding bounds to solve integral over domain.

Question

Let $\mathcal{D} = \lbrace (x,y)\in\mathbb{R}^{2}:1<xy<2,x^{2}<y<2x^{2}\rbrace.$ What is $$\iint_{\mathcal{D}} x^{3}+y^{3}\,\mathrm{d}y\,\mathrm{d}x \ \ ?$$

$(i)\ \ \frac{3}{4},$

$(ii) \ \ 1,$

$(iii) \ \ \frac{36}{37},$

$(iv) \ \ \frac{37}{36}.$

How can I find the bounds of my integrals here? The one for $\mathrm{d}y$ is evident it's $ x^2 $ to $ 2x^2,$ but what about the bounds on $x$ which has to be fixed in order to integrate?

Answer 1

Substitute $u=xy$ and $v=\frac{y}{x^2}$ so that the limits on both integrals are $[1,2]$.

On application of the Jacobian, the integrand is $$\frac{x^2}{3y}(x^3+y^3)=\frac 13(\frac{u}{v^2}+u^2)$$

The integrals are now very straightforward, and the answer you get is $\frac{37}{36}$ so the answer is (iv).

Answer 2

Try rewriting the inequalities like this: \begin{gather} \frac1x < y < \frac2x \\ x^2 < y < 2x^2 \end{gather}

Now how is it "evident" that the bounds of $dy$ are $x^2$ to $2x^2$? Why aren't they $\frac1x$ to $\frac2x$?

Look again at the graph of the region (credit: @DMcMor). A good first step is to find the "corners" of that region where the curves that form the different pieces of the boundary intersect. Then decide whether you want $dx$ or $dy$ to be the inner integral. If you want $dx$ to be the outer integral, as suggested in the question, find the points in the region that are farthest to the left and farthest to the right. Those will be lower and upper limits of $x,$ but not necessarily lower and upper limits of the same integral. Sometimes you need to "slice" a region into multiple pieces, cutting it along vertical lines, and do a separate $dx$ integral on each side of each line.