Double integral - calculating without translating a circle to $(0,0)$ So I'm a bit confused with calculating a double integral when a circle isn't centered on $(0,0)$. 
For example: Calculating $\iint(x+4y)\,dx\,dy$ of the area $D: x^2-6x+y^2-4y\le12$. 
So I kind of understand how to center the circle and solve this with polar coordinates. Since the circle equation is $(x-3)^2+(y-2)^2=25$, I can translate it to $(u+3)^2+(v+2)^2=25$ and go on from there.
However I would like to know if I could solve this without translating the circle to the origin. I thought I could, so I simply tried solving $\iint(x+4y)\,dx\,dy$ by doing this:
$\int_0^{2\pi}\,d\phi\int_0^5(r\cos\phi + 4r\sin\phi)r\,dr$ but this doesn't work. I'm sure I'm missing something, but why should it be different? the radius is between 0 and 5 in the original circle as well, etc.
So my questions are:


*

*How can I calculate something like the above integral without translating the circle to the origin? What am I doing wrong?

*I would appreciate a good explanation of what are the steps exactly when translating the circle. I kind of "winged it" with just saying "OK, I have to move the $X$ back by 3, so I'll call it $X+3$, the same with the $Y$ etc. If someone could give a clear breakdown of the steps that would be very nice :)
Thanks!
 A: \begin{align}
u & = x-3, \qquad du=dx,\qquad x=u+3, \\
v & = y-2, \qquad dv=dy, \qquad y = v+2, \\[6pt]
x+4y & = (u+3) + 4(v+2) = u+4v+11.
\end{align}
$$
\iint\limits_D (x+4y)\,dx\,dy = \iint\limits_{D\,-\,(3,2)} (u+4v+11)\,du\,dv.
$$
Since $D-(3,2)=\{(u,v):u^2+v^2\le 25\}$, you can then use polar coordinates with $r\cos\theta=u$ and $r\sin\theta=v$ so that $du\,dv=r\,dr\,d\theta$.
PS: If you insist on not translating the circle to the origin, you can do this:
$$
\int_{3-5}^{3+5} \left(\int_{2-\sqrt{25-(x-3)^2}}^{2+\sqrt{25-(x-3)^2}} (x+4y) \,dy \right)\,dx
$$
A: From the equation of circle
$$(x-3)^2+(y-2)^2=25$$
We get the formulas for the top half $y_2(x)$ and the bottom half $y_1(x)$ of the circle:
$$y_2(x)=2+\sqrt{5^2-(x-3)^2}=2+\sqrt{(8-x)(x+2)}$$
$$y_1(x)=2-\sqrt{5^2-(x-3)^2}=2-\sqrt{(8-x)(x+2)}$$
Where $-2\le x \le 8$.
So the integration becomes:
$$I=\iint\limits_D (x+4y)\,dx\,dy =\int_{-3}^{7}\left(\int_{y_1(x)}^{y_2(x)}(x+4y)\,dy \right)\,dx=\int_{-3}^{7}\left(2(8+x)\sqrt{(8-x)(x+2)}\right)\,dx=275\pi$$
The last integration here is given by Mathematica 7.0. You may use the standard technique (which I forgot) to integrate it by hand.
Thanks for correcting my mistake.
