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i've done some progress with this one but i get stucked right before the integration part:

$A)$ Use the right double integral and replacing $x=ar$cos$(\theta)$, $y=br$sin$(\theta)$ in order to find the area inside the curve: $$(\frac{x^2}{a^2}+\frac{y^2}{b^2})^2=\frac{x^2}{a^2}-\frac{y^2}{b^2}$$

  • note: assume $a,b > 0$

$b)$ Calculate the surface of the elipse: $(x-2y+3)^2+(3x+4y-1)^2=100$

I study by myself and try to solve this test for over a week so please give a full solution so i can study from,

Thanks in advance :)

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You say we have a curve given by the set of all points $(x,y) \in \mathbb{R}^2$ that satisfy:

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

This is not a curve. It's a horizontal line $y = 0$: $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{x^2}{a^2} - \frac{y^2}{b^2} \\ \frac{y^2}{b^2} = - \frac{y^2}{b^2} \\ 2\frac{y^2}{b^2} = 0 $$

(Since $y^2$ and $b^2$ are always 0, the only solution to this equation is $y=0$; $x$ is free to be anything.)

I don't have enough reputation yet to post a comment asking for clarification, so, I'll assume that you meant:

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

for some radius $R$.

Normally we might try switching the coordinate system from Cartesian coordinates to polar coordinates ($x = r\,cos(\theta)$, $y = r\,cos(\theta)$):

$$ \frac{(r\,cos(\theta))^2}{a^2} + \frac{(r\,sin(\theta))^2}{b^2} = \frac{(r\,cos(\theta))^2}{a^2} - \frac{(r\,sin(\theta))^2}{b^2} \\ $$

However, as the problem mentioned, the problem is not practical to solve unless you use your own coordinate system.

Let $r$ and $\theta$ be defined in such a way that:

$$ x = a\,r\,cos(\theta),\hspace{1em}y = b\,r\,sin(\theta), $$

(if you're curious, $r$ and $\theta$ are defined as

$$ r = \sqrt{\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2}, \theta = tan^{-1}\left(\frac{\left(\frac{y}{b}\right)}{\left(\frac{x}{a}\right)}\right) $$

but this is unimportant.)

The Jacobian determinant of this coordinate system conversion is:

$$ \begin{align*} \begin{vmatrix} J \end{vmatrix} &= \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\end{vmatrix} = \begin{vmatrix} a\,cos(\theta) & -a\,r\,sin(\theta) \\ b\,sin(\theta) & b\,r\,cos(\theta)\end{vmatrix}\\ &= a\,cos(\theta) \cdot b\,r\,cos(\theta) - (-a\,r\,sin(\theta) \cdot b\,sin(\theta)) \\ &= a\,b\,r\,cos^2(\theta) + a\,b\,r\,sin^2(\theta) \\ &= a\,b\,r \end{align*} $$

Since $dA$ is given by

$$ dA = \begin{vmatrix} J \end{vmatrix} \cdot dr \cdot dtheta $$

(you can refer to Paul's notes for a derivation of this), we have

$$ dA = a\,b\,r\,dr\,d\theta. $$

We can readily see that this reduces to the familiar $dA = r\,dr\,d\theta$ for the "regular" polar coordinate system where $a$ and $b$ each equal 1.

Anyway, now that we have the coordinate system conversion and its associated Jacobian determinant, we can go back:

$$ \frac{(a\,r\,cos(\theta))^2}{a^2} + \frac{(b\,r\,sin(\theta))^2}{b^2} = R^2 \\ r^2cos^2(\theta) + r^2sin^2(\theta) = R^2 \\ r = R \\ $$

We will integrate $dA$ over $\theta=0$ to $2\pi$, $r=0$ to $R$.

$$ A = \int_{0}^{2\pi} \int_{0}^{R} dA = \int_{0}^{2\pi} \int_{0}^{R} a\,b\,r\,dr\,d\theta \\ = \int_{0}^{2\pi} \frac{1}{2} \cdot a \cdot b \cdot (R^2 - 0^2) d\theta \\ = \frac{1}{2}\,a\,b\,(R^2 - 0^2)\,(2\pi - 0) $$

Now for your second question...

We are given an ellipse:

$$ (x - 2y + 3)^2 + (3x + 4y - 1)^2 = 100 $$

Once again, we want some coordinate system conversion from $x,y$ into $u,v$ such that we end up with:

$$ u^2 + v^2 = 100 $$

We can find the expressions for $u$ and $v$ in terms of $x$ and $y$, and vice versa, like this:

$$ u^2 = (x - 2y + 3)^2,\hspace{1em}v^2 = (3x + 4y - 1)^2 \\ \boxed{u = x - 2y + 3,\hspace{1em}v = 3x + 4y - 1} $$

$$ -3u = -3x + 6y - 9,\hspace{1em}v = 3x + 4y - 1 \\ -3u + v = -3x + 6y - 9 + 3x + 4y - 1 \\ 10y - 10 = -3u + v \\ \boxed{y = -\frac{3}{10}u + \frac{1}{10}v + 1} $$

$$ 2u = 2x - 4y + 6,\hspace{1em}v = 3x + 4y - 1 \\ 2u + v = 2x - 4y + 6 + 3x + 4y - 1 \\ 5x + 5 = 2u + v \\ \boxed{x = \frac{2}{5}u + \frac{1}{5}v - 1} $$

Now we are interested in the Jacobian determinant for converting the integral from $x,y$ to $u,v$, meaning, the one with partial derivatives of $x,y$ with respect to $u,v$. This is why I found $x$ and $y$ explicitly in terms of $u$ and $v$ for you. However, there's a trick. The multivariable inverse function theorem says that the Jacobian determinant from $x,y$ to $u,v$ is one divided by the Jacobian determinant from $u,v$ to $x,y$:

$$ \begin{vmatrix} J \end{vmatrix}_{u,v \rightarrow x,y} = \frac{1}{\begin{vmatrix} J \end{vmatrix}_{x,y \rightarrow u,v}}. $$

This means we did NOT need to derive $x$ and $y$ explicitly. Feel free to calculate the Jacobian determinant either way.

The Jacobian determinant for converting the integral from $x,y$ to $u,v$ is

$$ \begin{align*} \begin{vmatrix} J \end{vmatrix}_{x,y \rightarrow u,v} &= \frac{1}{\begin{vmatrix} J \end{vmatrix}_{u,v \rightarrow x,y}} = \frac{1}{\begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial v}{\partial x} \\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y}\end{vmatrix}} = \frac{1}{\begin{vmatrix} 1 & 3 \\ -2 & 4\end{vmatrix}} \\ &= \frac{1}{1 \cdot 4 - (3 \cdot -2)} = \frac{1}{4 + 6} = \frac{1}{10} \end{align*} $$

Now, the problem can be reworded like this: find the double integral $\iint\limits_D \frac{1}{10}dA$ inside the curve given by:

$$ u^2 + v^2 = 100 $$

This requires application of a second coordinate system conversion (on top of the one we just did) into polar coordinates, but that should be second nature.

In short:

$$ \text{Step 1:}\hspace{1em} x,y \rightarrow u,v \\ \text{Step 2:}\hspace{1em} u,v \rightarrow r,\theta \\ \text{Step 3:}\hspace{1em} \text{evaluate it, profit, et cetera} $$

$$ \iint\limits_D \frac{1}{10}dA \\ = \frac{1}{10} \int_{0}^{2\pi} \int_{0}^{\sqrt{100}} r\,dr\,d\theta \\ = \frac{1}{10} \int_{0}^{2\pi} \frac{1}{2} \cdot (\sqrt{100}^2 - 0^2) d\theta) \\ = \frac{1}{10} \int_{0}^{2\pi} \frac{1}{2} \cdot 100 d\theta) \\ = \frac{1}{10} \cdot \frac{1}{2} \cdot 100 \cdot 2\pi. $$

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  • $\begingroup$ Thank you for the second question, about the first i forgot to write something (i just corrected it) so it's not a line $\endgroup$ – user156698 Jun 12 '14 at 11:14

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