Double integral - domain of integration

Calculate the following double integral: $\int \int_S x\; dx \; dy$, where $S$ is the region limited by the circle of radius 1 and center (0,1) and the line that pass through the points (2,0) and (0,2)

I think that this integral must be

$$\int_0^1 dx \int_{-x+2}^{\sqrt{1-x^2}+1} x \; dy$$

but the answer is $\frac{1}{3}$ and this integral is $\frac{1}{6}$. What am i doing wrong?

• everything seems to be fine. – mookid Mar 28 '14 at 15:55 (after having spent some time working in the wrong region):

The statement of the boundaries is ambiguous, because they apply equally well to the blue or the green region shown in the graph above. But it wasn't a complete waste of time to integrate in the wrong place, as I will describe below.

The equation of the circle is $\ x^2 \ + (y-1)^2 \ = \ 1 \$ , so the functions for the "upper and lower" semicircles are $\ y \ = \ 1 \ \pm \ \sqrt{1-x^2} \ .$ The blue region is bounded above by the upper semi-circle and below by the line $\ x + y = 2 \$ . The intersection points of the line and circle are $\ (0,2) \ \text{and} \ (1,1) \ .$ So the double integral is

$$\int_0^1 \int^{1 + \sqrt{1-x^2}}_{2-x} \ x \ \ dy \ dx \ = \ \int_0^1 \ xy \ \vert^{1 + \sqrt{1-x^2}}_{2-x} \ \ dx$$

$$= \ \int_0^1 \ x \ \left[ \ (1 + \sqrt{1-x^2}) \ - \ (2-x) \ \right] \ \ dx \ = \ \int_0^1 \ x^2 \ - \ x \ + \ x \sqrt{1-x^2} \ \ dx$$

$$= \ \left[ \ \frac{1}{3}x^3 \ - \ \frac{1}{2}x^2 \ - \ \frac{1}{3} (1-x^2)^{3/2} \ \right] \ \vert_0^1$$

$$= \ \left( \ \frac{1}{3} \ - \ \frac{1}{2} \ - \ 0 \ \right) \ - \left( \ 0 \ - \ 0 \ - \ \frac{1}{3} \ \right) \ = \ \frac{1}{6} \ ,$$

which apparently does not agree with the "given answer". So who's right?

Let's take the integration over the green region, which is bounded below entirely by the lower semi-circle, bounded above by the upper semi-circle over $\ [-1 , 0 ] \$ , and above by the line over $\ [0 , 1] \ .$ So we have two integrals here:

$$\int_{-1}^0 \int^{1 + \sqrt{1-x^2}}_{1 - \sqrt{1-x^2}} \ x \ \ dy \ dx \ \ + \ \ \int_0^1 \int^{2-x}_{1 - \sqrt{1-x^2}} \ \ x \ \ dy \ dx$$

$$= \ \ \int_{-1}^0 \ \ x \ \left[ \ (1 + \sqrt{1-x^2}) \ - \ ({1 - \sqrt{1-x^2}}) \ \right] \ \ dx$$

$$+ \ \ \int_0^1 \ x \ \left[ \ (2-x) \ - \ (1 - \sqrt{1-x^2}) \ \right] \ dx$$

$$= \ \ 2 \ \int_{-1}^0 \ \ x \ \sqrt{1-x^2} \ \ dx \ + \ \ \int_0^1 \ x \ - \ x^2 \ + \ x \sqrt{1-x^2} \ \ dx$$

$$= \ \left[ \ - \ \frac{2}{3} (1-x^2)^{3/2} \ \right] \ \vert_{-1}^0 \ + \ \ \left[ \ \frac{1}{2}x^2 \ - \ \frac{1}{3}x^3 \ - \ \frac{1}{3} (1-x^2)^{3/2} \ \right] \ \vert_0^1$$

$$= \ \left[ \ -\frac{2}{3} \ + \ 0 \ \right] \ + \ \left[ \ \left( \ \frac{1}{2} \ - \ \frac{1}{3} \ - \ 0 \ \right) \ - \left( \ 0 \ - \ 0 \ - \ \frac{1}{3} \ \right) \ \right]$$

$$= \ -\frac{2}{3} \ + \ \frac{1}{6} \ + \ \frac{1}{3} \ = \ -\frac{1}{6} \ .$$

What was the point of doing this? Together, the green and blue regions cover the entire circle, which is symmetrical about the $\ y-$ axis. We are integrating $\ x \$ , a function with odd symmetry (about the $\ y-$ axis) over this region. So the integral of this function over the circle is zero. And indeed, the integrals over the blue and green regions have opposite signs.

So I believe the result of $\ \frac{1}{6} \$ for the blue region is reliable. My suspicion is that the solver may have made an error summing the fractions (I've seen it happen...).