Change of Variable Theorem on Riemann Integratable Function $A:=\{(x,y):x^2+y^2 \leq 1, y \geq 0\}$

Question

Let $A:=$$\{(x,y):x^2+y^2 \leq 1, y \geq 0\}$ and $f:A \to \mathbb R$ be a Riemann Integratable function.

$(i)$ For $\gamma > 0$, let $A_{\gamma}:= \{ (rcos(t),rsin(t)): r \in [\gamma,1], t \in [0, \pi] \} $.

Using the Change of Variable Theorem, show that if $f: A \to \mathbb R$ is Riemann integratable function then; $\int_{A_{\gamma}}f(x,y) d(x,y)= \int_{[\gamma,1]\times[0,\pi]}f(rcos(t),rsin(t))rd(r,t)$.

$(ii)$ Why doesn't the Change of Variable Theorem tell us (directly) that $\int_{A}f(x,y) d(x,y)= \int_{[0,1]\times[0,\pi]}f(rcos(t),rsin(t))rd(r,t)$?

I'm struggling to get my head round this one. Here are my ideas so far:

$(i)$ Convert $x$ and $y$ to polar coordinates; $x=rcos(t)=g(r,t)$ and $y=rsin(t)=g(r,t)$, where $0≤t≤π$ and $γ≤r≤1$.

$J = {\dfrac{\partial(x,y)}{\partial(r,t)}} = \begin{vmatrix} \dfrac{\partial x}{\partial r} & \dfrac{\partial x}{\partial t} \\ \dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial t} \end{vmatrix} = \begin{vmatrix} \cos(t) & \ -rsin(t) \\ \sin(t) & r\cos(t) \end{vmatrix} = rcos^2t+rsin^2t=r(cos^2t+sin^2t)=r $

So using the formula for change of variable involving the jacobian determinant:

$\int_{A_{\gamma}}f(x,y) d(x,y)= \int_{[\gamma,1]\times[0,\pi]}f(g(r,t),g(r,t))\|\frac{\partial(x,y)}{\partial(r,t)}|d(r,t)$

$=\int_{[\gamma,1]\times[0,\pi]}f(rcos(t),rsin(t))rd(r,t)$

For part $(ii)$ taking the polar coordinate conversion of $x$ and $y$ again, $\sqrt{x^2+y^2}=r$.

So then

$\begin{aligned} \iint_{A} \left( x^2 + y^2 \right) \text{ d}(x,y)& = \int_{0}^{\pi/4} \int_{0}^{1} \left( (r\cos(t))^2 + (r\sin(t))^2 \right) |r| \text{ d}(r,t) \\ & = \int_{0}^{\pi/4} \int_{0}^{1} r^2 |r| \text{ d}(r,t) \end{aligned}$

$=\begin{aligned}\int_{\ t=0}^{t=\pi}[(\frac{1}{4})dr]dt\end{aligned}=4\pi$

EDIT: Additional query, since I cannot use CoV directly, how would I prove that$ \int_{A}f(x,y) d(x,y)= \int_{[0,1]\times[0,\pi]}f(rcos(t),rsin(t))rd(r,t) $ ?

With the conditions $f:A \to \mathbb R$ is bounded and $f|_{A_n}$ is Riemann Integratable for every $n$. Let $A$ be a Jordan Measurable set and $(A_n)_{n=1}^{∞}$ a sequence of Jordan Measurable subsets of $A$ such that $v(A)=lim_{n \to ∞}v(A_n)$), I know how to prove that:

If $f$ is Riemann Integratable then $ \int_{A}f(x)dx=lim_{n \to ∞}\int_{A}f(x)dx$.

I considered using this and the knowledge from $(i)$, but I'm not sure how to tie them together.

Answer 1

You have the right idea for $(i)$, but there are two problems. The first is simply that you were asked to integrate a generic function $f(x,y)$, but your calculation is for a specific function $x^2 + y^2$. Why did you use this instead of $f$?

The second is the point of the this exercise, and is specifically pointed out by part $(ii)$. The CofV theorem does not directly apply to this case. You need to go back to the statement of that theorem, read the conditions under which it applies, and determine if they are satisfied by this problem. I don't know how this theorem is stated for you, but there is probably a requirement that the function be integrable on an open set containing the domain of integration.

Is that true here? If not, then you need to apply the theorem to smaller sets for which it is true, and figure out how to extend it to your problem.