Compute the following double integral

Question

Let $$I = \iint_D \frac{x^2+2x}{\sqrt{y}}\,dA,$$ where $D = \{y = \frac{1}{x}, y=\frac{2}{x}, x=y, y=\frac{x}{2}, x>0\}$

I graphed the functions from $D$, and the graph looked like this: graph So from this graph I can see that $D = D_1 \bigcup D_2$, where
$D_1 = \{(x,y)\mid 1 \leq x \leq 2, \frac{1}{x}\leq y \leq \frac{x}{2}\}$
$D_2=\{(x,y)\mid 1 \leq x \leq 2 , x \leq y \leq \frac{2}{x}\}$

So my integral would become: $$\int_1^2 \int_{\frac{1}{x}}^{\frac{x}{2}} \frac{x^2+2x}{\sqrt{y}} \, dy \, dx +\int_1^2 \int_x^{\frac{2}{x}}\frac{x^2+2x}{\sqrt{y}} \, dy \, dx. $$ Am I right?

Answer 1

If we take $x$ outer, first we find the smallest and biggest $x$ in the region. The smallest $x$ happens at the left corner, $(1,1)$, and the biggest happens at the right corner, $(2,1)$. To find the limits on $y$ we draw a curve of constant $x$, a vertical line, through the region. Near the left corner the line would intersect the boundary of the region at the hyperbola $y=1/x$ and the line $y=x$, but as the vertical line moves to the right things change for $x>\sqrt2$. After this point the line intersects the boundary of the region at $y=x/2$ and $y=2/x$, so for $\sqrt2<x<2$ the limits on $y$ are different so the $x$ integral has to be broken up into $2$ pieces: $$\int\int_D\frac{x^2+2x}{\sqrt y}d^2A=\int_1^{\sqrt2}\int_{1/x}^x\frac{x^2+2x}{\sqrt y}dy\,dx+\int_{\sqrt2}^2\int_{x/2}^{2/x}\frac{x^2+2x}{\sqrt y}dy\,dx$$ The observant reader will note that we were fortunate that the upper and lower corners both happened when $x=\sqrt2$; if the boundary $y=2/x$ were changed to $y=3/x$ for example, we would have had to integrate $3$ pieces.

If we instead took $y$ outer, the smallest $y$ happens at the lower corner, $(\sqrt2,1/\sqrt2)$ and the biggest $y$ at the upper corner, $(\sqrt2,\sqrt2)$. To find the limits on $x$ for a given $y$ we draw a curve of constant $y$, a horizontal line, through the figure. For $1/\sqrt2<y<1$ the line intersects the boundary of the region at $x=1/y$ and $x=2y$, but for $1<y<\sqrt2$, the line intersects the boundary of the region at $y=x$ and $y=2/x$. Again we were fortunate that the left and right corners happened at the same $y$. So this way we get $$\int\int_D\frac{x^2+2x}{\sqrt y}d^2A=\int_{1/\sqrt2}^1\int_{1/y}^{2y}\frac{x^2+2x}{\sqrt y}dx\,dy+\int_1^{\sqrt2}\int_y^{2/y}\frac{x^2+2x}{\sqrt y}dx\,dy$$ The integral can be done in one piece if we are allowed to use the Jacobian. Let $u=xy$, $v=x/y$. Then $x=(uv)^{1/2}$, $y=(u/v)^{1/2}$ and $$\begin{align}d^2A&=dx\,dy=\left|\det\begin{bmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{bmatrix}\right|du\,dv\\ &=\left|\det\begin{bmatrix}\frac12\left(\frac vu\right)^{1/2}&\frac1{2(uv)^{1/2}}\\\frac12\left(\frac uv\right)^{1/2}&-\frac12\left(\frac u{v^3}\right)^{1/2}\end{bmatrix}\right|du\,dv=\frac1{2v}du\,dv\end{align}$$ Now the region is bounded by the hyperbolas $xy=u=1$ and $xy=u=2$ and the lines $x/y=v=1$ and $x/y=v=2$, so it's rectangular in the $(u,v)$ coordinate system and we get $$\int\int_D\frac{x^2+2x}{\sqrt y}d^2A=\int_1^2\int_1^2\frac{uv+2(uv)^{1/2}}{\left(\frac uv\right)^{1/2}}\frac1{2v}dv\,du$$ All $3$ ways you should get $$\int\int_D\frac{x^2+2x}{\sqrt y}d^2A=-\frac{32}{21}2^{3/4}-\frac{272}{105}2^{1/4}+\frac{772}{105}$$

Answer 2

If we graph the region we see that we are bound by two regions which I will call $R_1$ and $R_2$. $R_1=\left\{\frac{1}{x}<y<x, 1\leq x\leq \sqrt2\right\}$ $R_2=\left\{x<y<\frac{2}{x}, \sqrt2\leq x\leq 2\right\}$ Let $f(x,y)=\dfrac{x^2+2x}{\sqrt{y}}$, the integral will be $$\int_{1}^{\sqrt2}\int_{\frac{1}{x}}^{x}f(x,y)\,\mathrm{d}y\,\mathrm{d}x+\int_{\sqrt2}^{2}\int_{x}^{\frac{2}{x}}f(x,y)\,\mathrm{d}y\,\mathrm{d}x$$

Answer 3

No, your equations for the two regions you picked are incorrect. For example, let's look at your region $D_1$: According to your definition, at $x=1.5$: $$ \frac{1}{1.5} \leq y \leq \frac{1.5}{2} \rightarrow \frac{2}{3}=0.33\leq y\leq 0.75 $$

Clearly, the graph of the region shows that at $x=1.5$, $y$ is about $>0.74$.

There are several ways to divide a region into sub-regions. Now, I am assuming that I understand correctly the regions you had in mind, which are the two regions separated by the line $x=1$. If that's the case, then, your first region (below the line $x=1$) should have been:

$$ D_1=\{(x,y)|1\leq x\leq \sqrt2, \frac{1}{x}\leq y\leq1\} \bigcup \{(x,y)|\sqrt{2}\leq x\leq 2, \frac{x}{2}\leq y\leq1\} $$

The number $\sqrt{2}$ is the $x$-intersection between the lines $y=\frac{1}{x}$ and $y=\frac{x}{2}$