Q: Bounds of integration in terms of variable being integrated I missed this problem on a quiz and I'm not understanding how to it. The problem is: $$\int_{-3}^0\int_0^{\sqrt{9-x^2}} 2x^2 + 2y^2 \, dxdy$$
In his notes online, my professor says the way to solve this is too switch the order of $dx$ and $dy$, but can you simply switch the order of these without changing the value of the integral? He doesn't specify anything besides simply switching the order of $dx$ and $dy$.

A: hint: Polar coordinates ! $x = r\cos t, y = r\sin t$. Can you figure out the rest....?

A: Just to add What @Deepsea has told, the polar coordinate version of the integral is
$$I = \int_{\pi/2}^{\pi} \int_{0}^{3} 2(r^3) dr d\theta$$
same thing as
$$I = \int_{0}^{\pi/2} \int_{0}^{3} 2(r^3) dr d\theta$$

A: This has got to be a typo! You're absolutely right in being suspicious of the limits of integration for $x$ depending on the same $x$ — that doesn't really make sense. So either the original order of integration was $\int_{-3}^0\int_0^{\sqrt{9-x^2}} (2x^2 + 2y^2)\,\color{red}{dy\,dx}$ or that limit of integration was $\int_{-3}^0\int_0^{\sqrt{9-\color{red}{y}^2}} (2x^2 + 2y^2)\,dx\,dy$.
Also suspicious is the claim (from your professor?) that "the way to solve it is to switch the order of $dx$ and $dy$". You may be given an exercise to switch the order of integration, just for the sake of practicing that skill. But the new integral won't be any easier — so it's a fine exercise, but it's not the way to solve it. As others have already pointed out, the best way to evaluate this integral is by switching to polar coordinates.
MORE TECHNICAL NOTE. It's not completely true that the limits of integration for $x$ cannot depend on $x$, as I said above myself. We certainly can set up something like $\int_0^\color{red}{x} \color{blue}{x}\,d\color{blue}{x}$. But it's still true that limits of integration for $x$ cannot depend on the same $x$. You see, in this integral the red $\color{red}{x}$ and the blue $\color{blue}{x}$ are two different things! They unfortunately are denoted by the same letter, thus causing a clash of variables (very bad style!), but the context distinguishes them as two totally different things.


*

*The blue $\color{blue}{x}$ is the variable of integration. It's a bound variable, or a dummy variable in a sense, since the result won't depend on it. And in fact, any letter can be used as the variable of integration in the same integral without changing its value. For example: $\int_1^2 2x\,dx=\int_1^2 2y\,dy=\int_1^2 2p\,dp=3$.

*The red $\color{red}{x}$ is more like a parameter. It will be present in the result, because we will be substituting it for the (dummy) variable of integration. For example: $\int_1^x 2x\,dx=\int_1^x 2y\,dy=\int_1^x 2p\,dp=x^2-1$.
So technically speaking, the double integral given in your post is meaningful. But evaluating it would result in an answer that depends on that parameter $x$ from the upper limit of integration. And I seriously doubt your Calculus professor actually meant that.
