Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $a,b$ be positive real numbers, and let $R$ be the region in $\Bbb R^2$ bounded by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Calculate the integral $$ \int\int_R\left(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)^{3/2}dx\,dy $$

my question is I don't know anything about $R$, the function $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is not the function of $R$, so then how can I get the answer? Could somebody give me some hints.

share|improve this question
    
Can you evaluate the integral when $a = b = 1$? –  Daniel Fischer Jan 19 at 19:18
3  
What do you mean "the region... is not the function of R"? You're given an ellipse and the integral is over that ellipse. –  DonAntonio Jan 19 at 19:18

3 Answers 3

\begin{equation*}I=\iint_{R}(1-\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}})^{3/2}\,dx\,dy\tag{1} \end{equation*}

my question is I don't know anything about $R$

The equation \begin{equation*} \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\tag{2} \end{equation*} represents an ellipse centered at $(x,y)=(0,0)$, with major and minor axes coinciding with the $x,y$ axes. This ellipse is the boundary of the region $R$.

\begin{equation*}R=\left\{ (x,y)\in\mathbb{R}^{2}:0\le \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\le 1\right\}\tag{3}\end{equation*}

the function $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is not the function of $R$

The integrand $(1-\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}})^{3/2}$ is evaluated over $R$.

To evaluate the integral $I$ we can proceed by making a transformation of the ellipse $(2)$ to a circle and then changing from cartesian to polar coordinates, or we could use right away polar coordinates.

  • If we make the change of variables $x=aX,y=bY$, then we get the circle centered at $(X,Y)=(0,0)$ and radius $1$ \begin{equation*}X^{2}+Y^{2}=1.\tag{4}\end{equation*}The region $R$ becomes the unit circle \begin{equation*}C=\left\{ (X,Y)\in\mathbb{R}^{2}:0\le X^2+Y^2\le 1\right\}\tag{5}\end{equation*} The Jacobian determinant of the transformation is just \begin{equation*}\left\vert \frac{\partial (x,y)}{\partial (X,Y)}\right\vert =ab.\tag{6}\end{equation*} This corresponds to the following linear relation between the area elements\begin{equation*}dx\,dy=ab\,dX\,dY.\tag{7}\end{equation*}As a consequence the given integral over $R$ can be rewritten as an integral over $C$ \begin{equation*}I=\iint_{C}(1-( X^{2}+Y^{2}) )^{3/2}\,ab\,dX\,dY.\tag{8}\end{equation*}
  • If we now use polar coordinates $r,\theta $ \begin{eqnarray*}X &=&r\cos \theta\\Y &=&r\sin \theta\\X^{2}+Y^{2}&=&r^{2}\tag{9}\end{eqnarray*} the integral $I$ becomes\begin{eqnarray*}I &=&ab\int_{r=0}^{1}\int_{\theta =0}^{2\pi }(1-r^{2})^{3/2}r\,dr\,d\theta\\ &=&2\pi ab\int_{0}^{1}(1-r^{2})^{3/2}r\,dr,\tag{10}\end{eqnarray*} because the Jacobian determinant \begin{equation*}\left\vert \frac{\partial (X,Y)}{\partial (r,\theta )}\right\vert =r.\tag{11}\end{equation*}In terms of area elements this means that they get converted by the relation \begin{equation*}dX\,dY=r\,dr\,d\theta .\tag{12}\end{equation*}
  • Finally, to evaluate $I$ we can make the substitution $t=r^{2}-1$ \begin{equation*}I=2\pi ab\int_{0}^{1}(1-r^{2})^{3/2}r\,dr=2\pi ab\int_{0}^{1}\frac{1}{2}t^{3/2}\,dt=\frac{2ab}{5}\pi.\tag{13}\end{equation*}
share|improve this answer
    
+1 for very complete source Americo.:) –  Babak S. Jan 20 at 15:46
    
@B.S. Thank you! –  Américo Tavares Jan 20 at 16:35

Using $$x=ar\cos(t), y=br\sin(t),~~ 0\leq t\leq 2\pi,~r\ge0$$ we get $$\int_{t=0}^{2\pi}\int_{r=0}^1(1-r^2)^{3/2}J(x,y)drdt$$ For $J(x,y)$ look at this similar one Example2.

share|improve this answer
    
@amWhy: Hello my friend Amy. Yes. As you know, I have to correct my claims in my recent article. Mathematics is a beautiful cruel one. –  Babak S. Jan 20 at 15:45

It sounds like you're just a bit confused about notation. $R$ is simply the name of the region. The notation

$$\iint\limits_{R} f(x,y) \, dA$$

simply means that we should integrate over the region $R$. In your case, $R$ is defined to be the region contained inside the ellipse

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$

For $a=3$ and $b=2$, this situation could be illustrated as follows:

enter image description here

As the other answers already indicate, the integral can be evaluated easily by a change of variables $x=ar\cos(\theta)$ and $y=br\sin(\theta)$. It can also be expressed as an iterated integral in Cartesian coordinates

$$4\int_0^3 \int_0^{\frac{b}{a}\sqrt{a^2-x^2}} \left(1-\frac{x^2}{a^2} - \frac{y^2}{b^2}\right) \, dy \, dx,$$

which evaluates to $2\pi a b/5$, though it's certainly more algebraically cumbersome than the change of variables approach.

share|improve this answer
    
+1, nice explanation. –  Américo Tavares Jan 20 at 0:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.