How to compute $\iint\frac {\mathrm dx\mathrm dy}{\sqrt {x^{2}-3y^{2}}}$, $\ x^{2}+y^{2}\leq 4x-3$?

Question

Question

$$\iint \dfrac {\mathrm dx\mathrm dy}{\sqrt {x^{2}-3y^{2}}},\quad x^{2}+y^{2}\leq 4x-3$$

Hint: use polar coordinates.

My Answer

$\left( x-2\right) ^{2}+y^{2}\leq 1$ then $C(2,0) , r=1$

$$\iint \dfrac {r\,\mathrm dr\mathrm d\theta }{r\sqrt {\cos ^{2}\theta -3\sin ^{2}\theta }}\Longrightarrow\int\dfrac {-\mathrm d\theta }{\sqrt {\left(1/2\right) ^{2}-\left( \cos \theta \right) ^{2}}}=-\sin^{-1} \left( 2\cos \theta \right)$$

boundary $0$ and $1$ for $r$ ? boundary $0$ and $2\pi$ for $\theta$?

Is my answer along with boundaries correct?

Answer 1

Yes the boundary are correct but we need to assume

• $x=2+r\cos \theta$

in order to have a correct parametrization for the circle.

Answer 2

Hint: Try with \begin{cases} x=2+r\cos\theta,\\ y=r\sin\theta. \end{cases} where $0\leq\theta\leq2\pi$, and $0\leq r\leq1$.