MHB Compute Area of 2 Simultaneous Inequalities: Don Leon at Yahoo Answers

[MarkFL]

Here is the question:

What is the area of the region defined by the following set of inequalities?


What is the area of the region defined by the following set of inequalities?
{ 0 < 2xy < 1 and 0< (x+y)/2 <1}

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Don Leon,

A simultaneous plot of the two inequalities:

$$0<2xy<1$$

$$0<\frac{x+y}{2}<1$$

is given here:

View attachment 1798

We need to know the $x$-coordinates of the intersections of the curves:

$$y=\frac{1}{2x}$$

$$y=2-x$$

We obtain the quadratic:

$$2x^2-4x+1=0$$

And application of the quadratic formula reveals:

$$x=1\pm\frac{1}{\sqrt{2}}$$

Hence, the area $A$ is given by:

$$A=\int_0^{1-\frac{1}{\sqrt{2}}} 2-x\,dx+\frac{1}{2}\int_{1-\frac{1}{\sqrt{2}}}^{1+\frac{1}{\sqrt{2}}}\frac{1}{x}\,dx+\int_{1+\frac{1}{\sqrt{2}}}^2 2-x\,dx$$

Application of the FTOC gives us:

$$A=\left[2x-\frac{1}{2}x^2 \right]_0^{1-\frac{1}{\sqrt{2}}}+\frac{1}{2}\left[\ln|x| \right]_{1-\frac{1}{\sqrt{2}}}^{1+\frac{1}{\sqrt{2}}}+\left[2x-\frac{1}{2}x^2 \right]_{1+\frac{1}{\sqrt{2}}}^2$$

Let's simplify one integral at a time:

$$\left[2x-\frac{1}{2}x^2 \right]_0^{1-\frac{1}{\sqrt{2}}}=2\left(1-\frac{1}{\sqrt{2}} \right)-\frac{1}{2}\left(1-\frac{1}{\sqrt{2}} \right)^2=2-\sqrt{2}-\frac{1}{2}\left(1-\sqrt{2}+\frac{1}{2} \right)=\frac{5}{4}-\frac{1}{\sqrt{2}}$$

$$\frac{1}{2}\left[\ln|x| \right]_{1-\frac{1}{\sqrt{2}}}^{1+\frac{1}{\sqrt{2}}}= \frac{1}{2}\ln\left(\frac{1+\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}} \right)= \frac{1}{2}\ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1} \right)=\ln\left(\sqrt{2}+1 \right)$$

$$\left[2x-\frac{1}{2}x^2 \right]_{1+\frac{1}{\sqrt{2}}}^2=\left(2(2)-\frac{1}{2}(2)^2 \right)-\left(2\left(1+\frac{1}{\sqrt{2}} \right)-\frac{1}{2}\left(1+\frac{1}{\sqrt{2}} \right)^2 \right)=$$

$$(4-2)-\left(2+\sqrt{2}-\frac{1}{2}\left(1+\sqrt{2}+\frac{1}{2} \right) \right)=\frac{3}{4}-\frac{1}{\sqrt{2}}$$

Adding the results, we obtain:

$$A=2-\sqrt{2}+\ln\left(\sqrt{2}+1 \right)$$

 

[MarkFL]

We could also use polar coordinates to find the area.

The line $$x+y=2$$ becomes $$r^2=\frac{4}{1+\sin(2\theta)}$$ and the curve $$2xy=1$$ becomes $$r^2=\csc(2\theta)$$.

Equating the two, we find they intersect at:

$$\theta=\frac{1}{2}\sin^{-1}\left(\frac{1}{3} \right)$$

And so the area $A$ may be expressed as:

$$A=4\int_0^{\frac{1}{2}\sin^{-1}\left(\frac{1}{3} \right)}\frac{1}{1+\sin(2\theta)}\,d\theta+ \int_{\frac{1}{2}\sin^{-1}\left(\frac{1}{3} \right)}^{\frac{\pi}{4}}\csc(2\theta)\,d\theta$$

Applying the FTOC, we find:

$$A=2\left[\tan\left(\theta-\frac{\pi}{4} \right) \right]_0^{\frac{1}{2}\sin^{-1}\left(\frac{1}{3} \right)}+\frac{1}{2}\left[\ln\left|\csc(\theta)-\cot(\theta) \right| \right]_{\sin^{-1}\left(\frac{1}{3} \right)}^{\frac{\pi}{2}}$$

After simplifying, we obtain:

$$A=2-\sqrt{2}+\ln\left(\sqrt{2}+1 \right)$$