MHB Calculating integral using polar coordinates

[mathmari]

Hey!

Using polar coordinates I want to calculate $\iint_D \frac{1}{(x^2+y^2)^2}dxdy$, where $D$ is the space that is determined by the inequalities $x+y\geq 1$ and $x^2+y^2\leq 1$.

We consider the function $T$ with $(x,y)=T(r,\theta)=(r\cos \theta, r\sin\theta)$.

From the inequality $x^2+y^2\leq 1$ we get that $r^2\leq 1 \Rightarrow -1\leq r\leq 1$. Since $r$ must be positive we get that $0\leq r\leq 1$.

How could we use the inequality $x+y\geq 1$ ? What do we get from $r\cos \theta+ r\sin\theta\geq 1$ ? (Wondering)

 

[MarkFL]

Suppose we set:

$$r\cos(\theta)+r\sin(\theta)=1$$

Square:

$$1+\sin(2\theta)=\frac{1}{r^2}$$

Also, consider from the other equation, we look at:

$$r^2=1$$

So, we then have:

$$\sin(2\theta)=0$$

And so we ultimately find:

$$0\le\theta\le\frac{\pi}{2}$$

Outer radius:

$$r_1=1$$

Inner radius:

$$r_2=\frac{1}{\sqrt{\sin(2\theta)+1}}$$

Hence:

$$\frac{1}{\sqrt{\sin(2\theta)+1}}\le r\le1$$

 

[I like Serena]

Alternatively, we can already see that $D$ is in the first quadrant, so $0\le \theta\le\frac \pi 2$, and:
$$r\cos\theta + r\sin\theta \ge 1 \quad\Rightarrow\quad
r\sqrt 2\sin(\theta+\frac\pi 4) \ge 1 \quad\Rightarrow\quad
r \ge \frac 1{\sqrt 2\sin(\theta+\frac\pi 4)}
$$

 

[mathmari]

We have that $0\leq \theta \leq \frac{\pi}{2}$.

Then from the inequality $x+y\geq 1$ we get the following: $$r\cos \theta+r\sin\theta\geq 1 \Rightarrow r(\cos \theta+\sin\theta)\geq 1$$
At this interval of $\theta$ we have that $\cos\theta\geq 0$ and $\sin\theta\geq 0$ and so $\cos\theta+\sin\theta\geq 0$.
So, we get $$r\geq \frac{1}{\cos \theta+\sin\theta}$$

From the inequality $x^2+y^2\leq 1$ we get that $r^2\leq 1\Rightarrow -1\leq r\leq 1$.

Therefore, we get $$\frac{1}{\cos \theta+\sin\theta}\leq r\leq 1$$

Your under bound is different. Have I done something wrong? (Wondering)

 

[I like Serena]

Your under bound is different. Have I done something wrong? (Wondering)

Nope. It's all correct. You have yet another expression for the same thing. (Nod)

 

[mathmari]

Nope. It's all correct. You have yet another expression for the same thing. (Nod)

Ah ok! Great! Thank you very much! (Smile)