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Given curve$$(a_1x+b_1y+c_1)^2 + (a_2x+b_2y+c_2)^2 = 1$$ where $$d=a_1b_2-a_2b_1 \neq 0$$
Find area $|D|$ under the curve.

Here's my try

Substitute $\left\{\begin{matrix} u := a_1x+b_1y+c_1\\ v := a_2x+b_2y+c_2 \end{matrix}\right.$

$$d = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1b_2-a_2b_1$$ $$d_x = \begin{vmatrix} u-c_1 & b_1 \\ v-c_2 & b_2 \end{vmatrix} = b_2(u-c_1) - b_1(v-c_2)$$ $$d_y = \begin{vmatrix} a_1 & u - c_1 \\ a_2 & v - c_2 \end{vmatrix} = a_1(v-c_2) - a_2(u-c_1)$$
$x= \frac{d_x}{d} = \frac{b_2(u-c_1)-b_1(v-c_2)}{d}$
$y= \frac{d_y}{d} = \frac{a_1(v-c_2)-a_2(u-c_1)}{d}$

The Jacobian $$|J| = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac{b_2}{d} & -\frac{b_1}{d}\\ -\frac{a_2}{d} & \frac{a_1}{d} \end{vmatrix} = \left | \frac{a_1b_2}{d} - \frac{a_2b_1}{d} \right | = \left |\frac{d}{d} \right | = 1$$

Now working with curve $u^2 + v^2 = 1$, we get $|D| = \iint dx dy =\iint |J| du dv = \iint du dv$
Substitute again $\left\{\begin{matrix} u=rcos\theta\\ v=rsin\theta \end{matrix}\right.$,$0 \leq r \leq 1$ and $0 \leq \theta \leq 2\pi$, $|J_2| = r$

I'm getting $\int_{0}^{2\pi} \int_{0}^{1} r d r d \theta = \pi$, but the answer is $\frac{\pi}{\left | d \right |}$. I think my calculations are correct, but my approach is wrong somewhere.

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    $\begingroup$ In the calculus of $|J|$ : the determinant gives you $d^2$ at the denominator. $\endgroup$ – Nicolas Nov 28 '15 at 13:51
  • $\begingroup$ Oh, I see, and I thought i calculated it correctly. thanks $\endgroup$ – shcolf Nov 28 '15 at 13:53
  • $\begingroup$ You're welcome! $\endgroup$ – Nicolas Nov 28 '15 at 13:53

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