I have to evaluate the integral $$\int\int_{R}2xdxdy$$ where $R$ is bounded by the curves $x^{2}+y^{2}=16$ and $x^{2}+y^{2}=4x$ in the first quadrant.

I have to use polar coordinates, so I get, for $x=r\cos\theta$ and $y=r\sin\theta$ the equations: $$r=4$$ and $$r=4\cos\theta$$ My point is: I know that, since I have to evaluate it in the 1st quadrant, so $0\leq\theta\leq\pi/2$, but about $r$? Is $4\cos\theta\leq r\leq4$? Cause, if so, I don't get the right answer I have here (it's said it's $16-3\pi$).


1 Answer 1


Yes, it is ok. The circle $x^2+y^2=4^2$ is the circle with center in $(0,0)$ and radius $4$. The other circle is $(x-2)^2+y^2=2^2$, centered in $(2,0)$ with radius $2$, it is passing through the origin, and it is tangent in $(4,0)$ to the first one.

The domain $R$ seems to be as follows, from the disk with boundary from the first circle we extract the disk with boundary the second circle, restrict then to the region in the first quadrant.

A ray with slope $\theta=t$ from the origin is then hitting the second circle in a point $A(t)$ such that $(0,0)$, and $(4,0)$, and $A(t)$ form a triangle with right angle in $A(t)$ and angle $t$ in the origin, so the segment from origin to $A(t)$ has length $4\cos t$. The integral is then $$ \iint_R2x\;dx\; dy = \int_0^{\pi/2}dt\int_{4\cos t}^4\; 2r\cos t\; r\; dr = \frac{128}3-8\pi\ . $$ The same is obtained if we integrate on the first quarter disk, subtracting the value from the second half disk. The integral on the half disk is simpler to visualize. We integrate there $2x=2(x-2)+4$. Because of the symmetry, integrating $2(x-2)$ is zero. And integrating $4$ is four times the area of the second half disk, i.e. $4\cdot \frac 12\pi\cdot 2^2$. This term explains the $-8\pi$. The integral on the quarter of the first disk $D$ can be done in polar coordinates, or using Fubini as $$ \iint_{D}2x\;dx\; dy = \int_0^42x\; dx\int_0^{\sqrt{4^2-x^2}}\; dy = \int_0^42x\; dx\;\sqrt{4^2-x^2} = -\frac 23(4^2-x^2)^{3/2}\Bigg|_0^4 = \frac{128}3\ . $$


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