Calculating a double integral on unit circle

Question

My question is: $\iint_\Omega xy\,dx\,dy$ where $\Omega$ is the first quadrant bounded by $x^2+y^2=1$

I've managed to integrate normally to give $\frac{x^2y^2}{2}+C+Cy$

How would I then apply the bounds of the integral?

Answer 1

I don't think you can integrate "normally" without first applying the bounds, since the bounds of the inter integral depend on the outer variable of integration.

Note that here, since you are in the first quadrant, both $x\ge 0$ and $y\ge 0$, so we can solve for either. Let's pick $y$: $$ y^2 = 1 - x^2 \text{ with } x,y \ge 0 \implies y = \sqrt{1-x^2}, $$ so we have $0 \le y \le \sqrt{1-x^2}$ and $0 \le x \le 1$.

Can you now apply those bounds and integrate correctly?

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UPDATE So we have $$ \begin{split} \int_0^1 \int_0^\sqrt{1-x^2} xy dydx &= \frac12 \int_0^1 x \left(1-x^2\right)dx \\ &= \frac{1}{-4} \int_0^1 (-2)x \left(1-x^2\right)dx \\ &= \frac{-1}{8} \left(1-x^2\right)^2 \\ &= \frac18. \end{split} $$

Answer 2

In first quadrant you have set $$\left\lbrace\begin{array}{} 0 \leqslant x \leqslant 1 \\ 0 \leqslant y \leqslant \sqrt{1-x^2} \end{array}\right\rbrace $$ So integral will be $$\int\limits_{0}^{1}\int\limits_{0}^{\sqrt{1-x^2}}f(x,y)\,dx\,dy$$ More easy it looks for so called polar coordinates $$\begin{cases} x=r \cos \phi \\ y= r \sin \phi \end{cases} $$ Area transfer to $$\left\lbrace\begin{array}{} 0 \leqslant \phi \leqslant \frac{\pi}{2} \\ 0 \leqslant r \leqslant 1 \end{array}\right\rbrace $$ and integral $$\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{1}rf(r \cos \phi,r \sin \phi)\,d\phi,dr$$

Answer 3

In polar coordinates $$\int_0^{\pi/2}\int_0^1( \sin \theta \cos \theta\;d\theta) (\;r\;dr \;) $$

separate integration..

$$\dfrac{1}{3}\cdot \dfrac{-\cos \theta}{2}|_0^{\pi/2} \quad= \dfrac{1}{3}\cdot -\dfrac12 \cdot -2=\dfrac{1}{3} $$