Finding Integral over a Region Bounded by 4 curves

Question

Hi there I tried solving the problem at the bottom, but I'm not quite sure how to do it. I attempted to draw it first and then came up with the following bounds of integration.

$$\int_{2x}^{x/4}\int_{1/x}^{4/x} e^{-xy/2} dydx$$

However, I couldn't compute this to a numerical value. Normally, I would use strictly numbers rather than variables for the bounds of the outermost integral, but if you sketch the curve, I'm pretty sure you wouldn't get the correct region if the outermost integral had numerical bounds. Is there any way to calculate this problem that I'm not seeing?

Let R be the region bounded by $y = 2x$, $y = \frac{x}{4}$, $y = \frac{4}{x}$, $y = \frac{1}{x}$.

Evaluate $\iint_R e^{-xy/2} dA$

Edit: I overlooked the fact that there are 2 regions bounded by these curves of equal area. I am only looking to integrate the bounded region in the first quadrant.

Answer 1

Here is a graph showing one of the two identical regions bounded by these graphs:

Where the red curve is $y=2x$, the blue is $y=\frac{x}{4}$, green is $y=\frac{4}{x}$, and yellow is $y=\frac{1}{x}$. To calculate the area, we can find the sum of the areas $A_1, A_2,$ and $A_3$, where $A_1$ is the area between the red and yellow curves from the intersection of the yellow and red curves to the intersection of the red and green curves, $A_2$ is the area between the green and yellow curves from the intersection between the red and green curves to the intersection between the yellow and blue curves, and $A_3$ is the area between the green and blue curves from the intersection between the blue and yellow curves to the intersection between the green and blue curves. Let us calculate these one at a time.

First we calculate $A_1$. To do this, we need the x-coordinates of the intersection points of the yellow and red curves and the red and green curves. These x-coordinates are $\frac{1}{\sqrt 2}$ and $\sqrt 2$ (you can verify this for yourself). Then $$A_1=\int_{\frac{1}{\sqrt 2}}^{\sqrt 2} (2x-\frac{1}{x})dx$$ When we evaluate the indefinite integral, we get $$\int (2x-\frac{1}{x})dx=x^2-\ln x+C$$ So $$A_1=(2-\ln (\sqrt 2))-(\frac{1}{2}-\ln (\frac{1}{\sqrt 2}))$$ $$A_1=2-\frac{1}{2}\ln (2)-\frac{1}{2}-\frac{1}{2}\ln (2)$$ $$A_1=\frac{3}{2}-\ln 2$$ Now we want to find $A_2$. The bounds for this area are $\sqrt 2$ and $2$, so $$A_2=\int_{\sqrt 2}^{2} (\frac{4}{x}-\frac{1}{x})dx$$ $$A_2=\int_{\sqrt 2}^{2} (\frac{3}{x})dx$$ This is a standard integral, and we can immediately say that $$A_2=3\ln(2)-3\ln(\sqrt 2)$$ $$A_2=\frac{3}{2}\ln2$$ Now we have one more area to find. The bounds for $A_3$ are $2$ and $4$, so $$A_3=\int_{2}^{4} (\frac{4}{x}-\frac{x}{4})dx$$ And the indefinite integral is equal to $$\int (\frac{4}{x}-\frac{x}{4})dx=4\ln x -\frac{x^2}{8}+C$$ So $$A_3=(4\ln 4 -\frac{4^2}{8})-(4\ln 2 -\frac{2^2}{8})$$ $$A_3=8\ln 2 -\frac{16}{8}-4\ln 2 +\frac{4}{8}$$ $$A_3=4\ln 2 -\frac{3}{2}$$ Now we have all three areas, and we need only find their sum. The entire area $A$ is $$A=A_1+A_2+A_3$$ $$A=\frac{3}{2}-\ln 2+\frac{3}{2}\ln2+4\ln 2 -\frac{3}{2}$$ $$A=\frac{3}{2}\ln2+3\ln 2$$ $$A=\frac{9}{2}\ln2$$ Which should be the final answer. Is this correct?

Answer 2

To evaluate this integral most efficiently, it's better to switch to new variables that better reflect the region of integration (and won't ruin the integrand). Here's an outline of the solution.

Let $u=\frac{y}{x}$. Then two of the bounding equations, $y=\frac{x}{4}$ and $y=2x$ tell us that $u$ ranges in the interval $u\in\left[\frac{1}{4},2\right]$.

Let $v=xy$. Then the other two of the bounding equations, $y=\frac{1}{x}$ and $y=\frac{4}{x}$ tell us that $v$ ranges in the interval $v\in[1,4]$.

Now the given double integral can be rewritten as $$\iint_R e^{xy/2}\,dA=\int_{1/4}^2\int_1^4 e^{-v/2}\,|J|\,dv\,du,$$ where $J$ is the Jacobian of this change of variables.

Solving $u=\frac{y}{x}$ and $v=xy$ for $x$ and $y$, we find that $x=u^{-1/2}v^{1/2}$ and $y=u^{1/2}v^{1/2}$. Therefore the Jacobian is $$J=\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}=\begin{vmatrix} -\frac{1}{2}u^{-3/2}v^{1/2} & \frac{1}{2}u^{-1/2}v^{-1/2} \\ \frac{1}{2}u^{-1/2}v^{1/2} & \frac{1}{2}u^{1/2}v^{-1/2} \end{vmatrix}=-\frac{1}{2}u^{-1}.$$

Then the integral can be computed further as $$\int_{1/4}^2\int_1^4 e^{-v/2}\,|J|\,dv\,du=\int_{1/4}^2\int_1^4 e^{-v/2}\cdot\frac{1}{2}u^{-1}\,dv\,du=\frac{1}{2}\cdot\int_{1/4}^2 u^{-1}\,du\cdot\int_1^4 e^{-v/2}\,dv \\ =\frac{1}{2}\cdot\left[\ln|u|\right]_{1/4}^2\cdot\left[-2e^{-v/2}\right]_1^4=-\left(\ln2-\ln\frac{1}{4}\right)\cdot\left(e^{-2}-e^{-1/2}\right)=\left(e^{-1/2}-e^{-2}\right)\ln8.$$