Evaluate double integral $∬xydxdy$ Help me evaluate integral:
$$\iint xy\ dx\ dy$$
And Area:
$$x^2 + y^2 \leq 25 \quad 3x + y \geq 5$$
 I try find it in polar coordinates, and i think
$$ 0 \leq r \leq 5$$ but $\phi$ ? 
 A: The line $3x+y=5$ intersects the circle centered at $(0,0)$ and radius $5$ at two points: $(3,-4)$ and $(0,5)$. So, $\phi$ goes from $\arctan\left(-\frac43\right)$ to $\frac\pi2$.
A: That line, in polar coordinates, is $3r\cos(\theta)+r\sin(\theta)\geq5$. That is the same as $$r=\frac{5}{3\cos\theta+\sin\theta}$$
In polar, you have 
$$\int_{\theta_1}^{\theta_2}\int_{r=\frac{5}{3\cos\theta+\sin\theta}}^5r\cos(\theta)r\sin(\theta)r\,dr\,d\theta=\int_{\theta_1}^{\theta_2}\int_{r=\frac{5}{3\cos\theta+\sin\theta}}^5\frac{1}{2}\sin(2\theta)r^3\,dr\,d\theta$$
where you need to work out $\theta_1$ and $\theta_2$. (I see that has been done elsewhere in a comment.)
A: Given the straight line boundary off center, there is no advantage using the polar coordinates. Instead, integrate as follows,
$$A=\int_{-4}^5 \int_{\frac{5-y}3}^\sqrt{25-y^2}
xydxdy$$
where the lower and upper limits -4 and 5 are the $y$-intersections of  the line with the circle. The $x$-integration leads to
$$A=\frac12 \int_{-4}^5 y \left( 25-y^2-\frac{(5-y)^2}9\right)
dy=\frac{135}4$$
A: If $y=5-3x$ then $x^2+y^2=x^2+25-30x+9x^2=25$ so $10x(x-3)=0$. The points of intersection are therefore $(x,y)\in\left\{(0,5),(3,-4)\right\}$. The easiest way to do this integral seems to be $x$ inner, $y$ outer:
$$\begin{align}\int\int_Rxy\,d^2A&=\int_{-4}^5\int_{(5-y)/3}^{\sqrt{25-y^2}}xy\,dx\,dy=\int_{-4}^5\left[\frac12x^2\right]_{(5-y)/3}^{\sqrt{25-y^2}}y\,dy\\
&=\frac12\int_{-4}^5\left[25-y^2-\frac19\left(25-10y+y^2\right)\right]y\,dy\\
&=\frac1{18}\int_{-4}^5\left[200+10y-10y^2\right]y\,dy\\
&=\frac59\left[10y^2+\frac13y^3-\frac14y^4\right]_{-4}^5=\frac{135}4\end{align}$$
