# Multivariable integral with unclear bounds

1. Aug 22, 2016

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! I'm trying to solve the following problem and I am unsure about what I did:

Calculate $\int_M f(x,y) dx dy$ with $M = \{ (x,y) \in \mathbb{R}^2: x^2 + y^2 \leq 1, x \geq 0, y \geq 0 \}$ and $f(x,y) = xy^2$.

2. Relevant equations

One equation I'd like to discuss comes from the script of my teacher and is:

$\int_{x^2 + y^2 \leq R^2} f(x,y) dx dy = \int_0^R r \bigg( \int_0^{2π} f(r \cos \phi, r \sin \phi) d \phi \bigg) dr$

He claims that works when $f: \mathbb{R}^2 \to \mathbb{R}$ is continuous and $R > 0$. In general the conversion to polar coordinates is no problem for me, but here I'm wondering where does the first $r$ on the right-hand side come from?

3. The attempt at a solution

So at first I tried to use the equation given above without too much thinking, and I repeatedly came up with $0$ as a result. Then I tried to derive the bounds from the conditions present in $M$, and here's my go:

$x^2 + y^2 \leq 1 \implies r^2 (\cos^2 \phi + \sin^2 \phi) \leq 1 \implies 0 \leq r \leq 1$

$r \cos \phi \geq 0 \implies \cos \phi \geq 0 \implies -\frac{\pi}{2} \leq \phi \leq \frac{\pi}{2}$
$r \sin \phi \geq 0 \implies \sin \phi \geq 0 \implies 0 \leq \phi \leq \pi$
$\implies 0 \leq \phi \leq \frac{\pi}{2}$

Is that correct? Then I solved the integral with the bounds that I found:

$\int_M f(x,y) dx dy = \int_0^{\pi/2} \bigg( \int_0^1 r^4 \cos \phi \sin^2 \phi\ dr \bigg) d\phi$
$= \int_0^{\pi/2} \bigg[ \frac{r^5}{5} \cos \phi \sin^2 \phi \bigg]_0^1 d\phi$
$= \int_0^{\pi/2} \frac{1}{5} \cos \phi \sin^2 \phi\ d\phi$
$= \frac{1}{5} \bigg[ \frac{\sin^3 \phi}{3} \bigg]_0^{\pi/2}$ (I skipped the substitution part)
$= \frac{1}{15}$

As you can see, I put the $r$ of the equation above inside the integral, though I am not sure why it is there. Is that the correct way to tackle the problem in general?

Julien.

2. Aug 22, 2016

### Staff: Mentor

If I understand your question correctly, the "extra" r comes from conversion from an iterated integral in Cartesian (or rectangular) form, to polar form. In general, an integral of the form $\int \int f(x, y) dx dy$ becomes $\int \int g(r, \theta) r~drd\theta$.

3. Aug 22, 2016

### JulienB

Hi @Mark44 and thanks for your answer. Yes that was my first question, I can see now where this $r$ comes from. :)

Julien.

4. Aug 22, 2016

### Ray Vickson

Have you really never seen the change-of-variable formulas for double or triple integrals? If not, you should study that material; it is entirely standard, and of great importance in future studies; it also can save you hours of work.

See, eg.,
http://tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx
or
http://math.oregonstate.edu/home/pr...ulusQuestStudyGuides/vcalc/change/change.html

5. Aug 23, 2016

### JulienB

Hi @Ray Vickson! No I had seen them, something in my head didn't connect when seeing that formula in the middle of the script of my teacher. But what about the integral? Did I do it correctly, especially the bounds bit?

Thanks for the links. :)

Julien.

6. Aug 23, 2016

### SammyS

Staff Emeritus
Yes. It looks correct to me .

7. Aug 23, 2016

### JulienB

@SammyS That's great, thanks a lot for reviewing my work!

Julien.