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How can I find the area of the following region which is enclosed by the following curve:

$$(x+y)^2+(x+3y)^2=1$$

This is an ellipse, and I graphed it to find that its center is at the origin. Not sure where this leads to though. I know that the area comes out to $\frac{\pi}{2}$ from the back of the book, but not sure how to do this.

Shauna Rubinstien
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1 Answers1

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For a double integral approach, take $u=x+y$ and $v=x+3y$ so that $$\vec r=\frac{1}{2}(3u-v)\vec i+\frac{1}{2}(v-u)\vec j+0\vec k$$ Now, compute the Jacobian:

$$\vert \vec r_{u}\times \vec r_{v}\vert =\vert (\frac{3}{2}\vec i-\frac{1}{2}\vec j)\times (-\frac{1}{2}\vec i+ +\frac{1}{2}\vec j)\vert =\frac{1}{2}$$.

The area is then $\frac{1}{2}\int \int _{D}dudv=\frac{1}{2}\pi$ because $D$ is just the unit circle.

Matematleta
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  • May I ask how you got that parameterisation $\vec{r}(u,v)$? – Khallil Jul 27 '15 at 20:34
  • Can you add more steps to how you computed the Jacobian? Because I'm not familar to taking a Jacobian of vectors. – Shauna Rubinstien Jul 27 '15 at 20:36
  • You know that the point $(x,y,z)\in \mathbb R^{3}$ is given by $\vec r=x\vec i+y\vec j+z\vec k$ in rectangular coordinates. With $z=0$ and $u $ and $v$ defined as in the answer, when you solve for $x$ and $y$ in terms of $u $ and $v$, you get $\vec r$ as advertised. I chose the parameterization becuase it makes the integral trivial. – Matematleta Jul 27 '15 at 20:38
  • @ShaunaRubinstien: Sure. If you do not want to express this vectorially, then you write $u=x+y$ and $v=x+3y$, from which $x=\frac{1}{2}(3u-v)$ and $y=\frac{1}{2}(v-u)$. Then, you compute the four partial derivatives that give you the matrix elements, whose determinant is the jacobian. – Matematleta Jul 27 '15 at 20:42
  • @Chilango Can you please solve it this way? – Shauna Rubinstien Jul 27 '15 at 20:43
  • @ShaunaRubinstien:$ \vert \det \mathcal J\vert = \left | \frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u} \right |=\frac{1}{2}$ – Matematleta Jul 27 '15 at 20:51
  • @Chilango I get $\frac{-uv-1}{2}$ What am I doing wrong? For the partial derivatives I get, $(\partial x/ \partial u, \partial y/ \partial u, \partial x/ \partial v, \partial y/ \partial v)=(1/2(3-v), 1/2(v-1), 1/2(3u-1), 1/2(1-u))$ Can you find where I'm going wrong? – Shauna Rubinstien Jul 27 '15 at 20:56
  • @ShaunaRubinstien: I think your derivatives are wrong. Remember, when you take the partial wrt $u$, then $v$ is a constant. Likewise, when you take the partial wrt $v$, then $u$ is a constant. so for example, $\frac{\partial x}{\partial u}=\frac{1}{2}(3-0)$ – Matematleta Jul 27 '15 at 21:02