strange set of domain for integration

Question

Draw the plane $$ D = \dfrac{x}{4} < y < x, \quad xy < 16 $$ and describe it as a horizontally simple set and as a vertically simple set. Then calculate the integral $$ \int\int_D \mathrm{e}^{xy} \cdot y^2 \,\mathrm{d}x \,\mathrm{d}y. $$

Answer 1

The region you want is above the line $y=x/4$, below the line $y=x$, and below the curve $y=16/x$. You need to find the intersection points of these three curves.

• $y=x/4$ and $y=x$ intersect when $x/4=x$, so $(3/4)x=0$ and thus $x=0$. So this intersection point is $(0,0)$.
• $y=x/4$ and $y=16/x$ intersect when $x/4=16/x$, so $x^2=64$. Thus $x=\pm 8$ and $y=\pm 2$, so the intersection points are $(8,2)$ and $(-8,-2)$.
• $y=x$ and $y=16/x$ intersect when $x=16/x$, so $x^2=16$. Thus $x= \pm 4$ and $y = \pm 4$. So the intersection points are $(4,4)$ and $(-4,-4)$.

Now if you plot these lines and curves (look at the third plot) and shade the regions where the inequalities $x/4<y<x$ and $xy<16$ hold, you'll see that the three intersection points you need are $(0,0),(4,4)$, and $(8,2)$.

To set up the integral, you need to decide whether you want $dxdy$ or $dydx$. It seems more difficult to integrate $e^{xy}y^2$ with respect to $y$ than with respect to $x$, so I would do $dxdy$. (Either way you will need to split the integral into two integrals.)

To find the bounds of integration, we need to find constant bounds on $y$. Looking at the sketch of the region and the points we found, $y=0$ is the lower bound and $y=4$ is the upper bound. So we have $$\int_0^4 \int_?^? e^{xy}y^2 dxdy.$$ Now we need to find bounds on $x$ in terms of $y$. To find the lower bound on $x$, we need a curve on the left boundary of the domain. Looking at the sketch, that will be the curve $y=x$, which we can rewrite as $x=y$. This gives $$\int_0^4 \int_y^? e^{xy}y^2 dxdy.$$ To find the upper bound on $x$, we need a curve on the right boundary of the domain. We see, though, that two different curves bound the right edge of the domain. Sometimes it's $y=x/4$, and sometimes it's $y=16/x$. To determine where the change occurs, we need to know the intersection point. We already found that this point is $(8,2)$. So we need to split the integral at $y=2$ and write $$\int_0^2 \int_y^? e^{xy}y^2 dxdy+\int_2^4 \int_y^? e^{xy}y^2 dxdy.$$ From $y=0$ to $y=2$, the upper bound on $x$ is $y=x/4$, or $x=4y$; and from $y=2$ to $y=4$, the upper bound on $x$ is $y=16/x$, or $x=16/y$. This gives the final answer as $$\int_0^2 \int_y^4y e^{xy}y^2 dxdy+\int_2^4 \int_y^{16/y} e^{xy}y^2 dxdy.$$ I want to stress the importance of drawing a good sketch of the domain and labeling points of intersection when doing a problem like this. It's extremely difficult to solve a problem like this without visualizing the domain.

I'm not sure what is meant by "horizontally simple" and "vertically simple" sets.