Proving Inequality: Double Integral (dA / (4+x^2+y^2)) ≤ π

[PsychonautQQ]

Homework Statement

Prove the inequality double integral (dA / (4+x^2+y^2)) is less than or equal to pi, where the double integral has a sub D where D is the disk x^2 + y^2 less than or equal to four

Homework Equations

The Attempt at a Solution

I really have no idea, anyone want to give me a clue to help me get started?

 

[dirk_mec1]

Are you sure this is the exact question?

 

[Mark44]

Homework Statement

Prove the inequality double integral (dA / (4+x^2+y^2)) is less than or equal to pi, where the double integral has a sub D where D is the disk x^2 + y^2 less than or equal to four

Homework Equations

The Attempt at a Solution

I really have no idea, anyone want to give me a clue to help me get started?

Are you sure this is the exact question?

I believe it is.

Here's the integral and inequality:
$$\int_D \frac{dA}{4 + x^2 + y^2} \leq \pi$$
where D is the disk x² + y² ≤ 4.

The key here, I believe, is that $\frac{1}{4 + x^2 + y^2} \leq \frac 1 4$.

 

[dirk_mec1]

I believe it is.

Are you really sure?

 

[Office_Shredder]

Are you really sure?

Obvously only Psychonaut can be sure but the problem statement is a true statement if that's what you're trying to get at.

 

[dirk_mec1]

What do you mean by a "true statement"?

\int_0^4 \frac{r}{4+r^2}\ \mbox{d}r = \ln(\sqrt5) = 0.8

 

[Mark44]

Obvously only Psychonaut can be sure but the problem statement is a true statement if that's what you're trying to get at.

What do you mean by a "true statement"?

By "true statement" I think Office_Shredder means that the problem as described in the OP represents a problem that can be solved. In this case, the problem is fairly simple. If I'm missing something, please enlighten me.

 

[D H]

What do you mean by a "true statement"?

\int_0^4 \frac{r}{4+r^2}\ \mbox{d}r = \ln(\sqrt5) = 0.8

You have the wrong integration limits here. They should be from 0 to 2, not from 0 to 4.