Double integral $\iint_D |x^3 y^3|\, \mathrm{d}x \mathrm{d}y$

Question

Solve the following double integral \begin{equation} \iint_D |x^3 y^3|\, \mathrm{d}x \mathrm{d}y \end{equation} where $D: \{(x,y)\mid x^2+y^2\leq y \}$.

Some help please? Thank you very much.

Answer 1

Hints:

$$x^2+y^2\le y\iff x^2+\left(y-\frac12\right)^2\le\frac14$$

so the integration region is the disk of radius $\,0.5\,$ and center at $\,(0,0.5)\,$ . It's easy to see that within this region we always have $\,y\ge 0\;$ , and $\,x\,$ is positive dependeing on which half disk we're on: the right ($\,D^+\,$) or the left ($\,D^-\;$) one, so

$$\int\int\limits_D\left|x^3y^3\right|dxdy=\int\int\limits_{D^+} x^3y^3dxdy-\int\int\limits_{D^-}x^3y^3dxdy$$

So for example let's try to work with

$$\int\int\limits_{D^+}x^3y^3dxdy=\int\limits_0^1\int\limits_0^{\sqrt{\frac14-\left(y-\frac12\right)^2}}y^3x^3dxdy=\frac14\int\limits_0^1y^3\left(y-y^2\right)^2dy=$$

$$=\frac14\left(\frac16-\frac27+\frac18\right)=\frac1{672}\;\;\ldots\ldots$$

Answer 2

The following area illustrates $D$:

Obviously, $y\ge0$ and $x\ge 0$ while $0\le\theta<{\pi/2}$ and $x<0$ while ${\pi/2}\le\theta\le \pi$. So, it seems that the polar coordinates are the best measurement here. The following integrals are our new integrals to solve:

$$\int_{\theta=0}^{\pi/2}\int_{r=0}^{\sin\theta}(r^3\cos^3\theta\times r^3\sin^3\theta)r dr d\theta-\int_{\theta={\pi/2}}^{\pi}\int_{r=1}^{\sin\theta}(r^3\cos^3\theta\times r^3\sin^3\theta)r dr d\theta$$

Answer 3

This may be done easily in polar coordinates; the equation of the circle is $r=\sin{\theta}$, $\theta \in [0,\pi]$. The integrand is $r^6 |\cos^3{\theta} \sin^3{\theta}|$, and is symmetric about $\theta = \pi/2$. The integral is then

$$2 \int_0^{\pi/2} d\theta \, \cos^3{\theta} \, \sin^3{\theta}\, \int_0^{\sin{\theta}} dr \, r^7 = \frac14 \int_0^{\pi/2} d\theta\, \cos^3{\theta} \, \sin^{11}{\theta} $$

which may be evaluated simply as

$$ \frac14 \int_0^{\pi/2} d(\sin{\theta}) (1-\sin^2{\theta}) \sin^{11}{\theta} = \frac14 \left (\frac{1}{12} - \frac{1}{14} \right ) = \frac{1}{336} $$

Answer 4

Hint \begin{align} \mathop{\int\!\!\int}_{\substack{x^2+y^2\leq y}} |x^3y^3| \; \mbox{d} x\, \mbox{d} y = & \quad 2\!\!\!\!\!\!\!\!\!\!\mathop{\int\!\!\int}_{\begin{array}{rcl}x^2+(y-0.5)^2\!\!\!\!\!&\leq&\!\!\!\!\! 0.5^2\\ 0\leq x\!\!\!\!\!&\leq&\!\!\!\!\! 0.5\end{array}} x^3y^3 \; \mbox{d} x\, \mbox{d} y \\ = & \quad 2\!\!\!\!\!\!\!\!\!\!\mathop{\int\!\!\int}_{\begin{array}{rcl}0\leq y\!\!\!\!\!&\leq&\!\!\!\!\! +0.5+\sqrt{0.5^2-x^2}\\ 0\leq x\!\!\!\!\!&\leq&\!\!\!\!\! 0.5\end{array}} x^3y^3 \; \mbox{d} x\, \mbox{d} y \\ \end{align}

Answer 5

Since $$ D=\{(x,y)|\ x^2+y^2\le y\}=\left\{(x,y)|\ x^2+(y-\frac12)^2\le \frac14\right\}, $$ we have $y\in [0,1]$ for all $(x,y) \in D$.

Setting $$ x=r\cos\theta,\ y=r\sin\theta+\frac12,\quad \theta \in [-\pi/2,3\pi/2],\ r \in [0,\frac12], $$ we get \begin{eqnarray} \int_D|x^3y^3|\,dxdy&=&\int_Dy^3|x|^3\,dxdy=\int_{-\pi/2}^{3\pi/2}\int_0^{\frac12}r^4|\cos^3\theta|(r\sin\theta+\frac12)^3\,drd\theta\\ &=&\int_0^{\frac12}\int_{-\pi/2}^{\pi/2}r^4\cos^3\theta(r\sin\theta+\frac12)^3\,drd\theta\\ &&-\int_0^{\frac12}\int_{\pi/2}^{3\pi/2}r^4\cos^3\theta(r\sin\theta+\frac12)^3\,drd\theta\\ &=&\int_0^{\frac12}\int_{-\pi/2}^{\pi/2}r^4\cos^3\theta(r\sin\theta+\frac12)^3\,drd\theta\\ &&-\int_0^{\frac12}\int_{-\pi/2}^{\pi/2}r^4\cos^3\theta(r\sin\theta-\frac12)^3\,drd\theta\\ &=&2\int_0^{\frac12}\int_0^{\pi/2}r^4\cos^3\theta\left[(r\sin\theta+\frac12)^3-(r\sin\theta-\frac12)^3\right]\,drd\theta\\ &=&2\int_0^{\frac12}\int_0^{\pi/2}r^4\cos^3\theta(3r^2\sin^2\theta+\frac14)\,drd\theta\\ &=&\int_0^1(\frac{3t^2}{7\cdot2^6}+\frac{1}{320})(1-t^2)\,dt\\ &=&\int_0^1\left[\frac{3}{7\cdot2^6}(t^2-t^4)+\frac{1}{320}(1-t^2)\right]\,dt\\ &=&\frac{3}{7\cdot2^6}(\frac13-\frac15)+\frac{1}{320}(1-\frac13)=\frac{1}{336}. \end{eqnarray}