Question

If G is the region in \(\R^2\) given by
\(G=\left\{(x,y)\in \R^2:x^2+y^2<1,\frac{x}{\sqrt3}< y < \sqrt3x,x>0,y>0\right\}\)
then the value of
\(\frac{200}{\pi}\iint_Gx^2dx\ dy\)
is equal to _________. (Rounded off to two decimal places)

Answer 1

Correct Answer: 4.15 - 4.17

To solve the problem, we must evaluate the double integral \(\iint_G x^2 \, dx \, dy\) over the region \(G\). The region is defined by the inequalities:
\(x^2 + y^2 < 1, \frac{x}{\sqrt{3}} < y < \sqrt{3}x, x > 0, y > 0\).

Convert to polar coordinates where \(x = r \cos \theta\) and \(y = r \sin \theta\). The Jacobian of this transformation is \(r\), so \(dx \, dy = r \, dr \, d\theta\).

The inequality \(x^2 + y^2 < 1\) becomes \(r^2 < 1\), or \(r < 1\). The angular bounds are determined by \(\frac{x}{\sqrt{3}} < y < \sqrt{3}x\).

Translate \(\frac{x}{\sqrt{3}} < y\) to \(r \cos \theta/\sqrt{3} < r \sin \theta\), simplifying to \(\tan \theta > \frac{1}{\sqrt{3}}\), then \(\theta > \frac{\pi}{6}\). Similarly, \(\tan \theta < \sqrt{3}\) gives \(\theta < \frac{\pi}{3}\).

Thus, the region in polar coordinates is:
\(\frac{\pi}{6} < \theta < \frac{\pi}{3}, \, 0 < r < 1\).

The integral becomes:
\(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \int_0^1 (r \cos \theta)^2 \cdot r \, dr \, d\theta = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \int_0^1 r^3 \cos^2 \theta \, dr \, d\theta\)

\(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\) so the inner integral is:
\(\int_0^1 r^3 \frac{1 + \cos 2\theta}{2} \, dr = \left[ \frac{r^4}{8} + \frac{r^4 \cos 2\theta}{8} \right]_0^1 = \frac{1}{8} + \frac{\cos 2\theta}{8}\).

The outer integral then is:
\(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left(\frac{1}{8} + \frac{\cos 2\theta}{8}\right) \, d\theta = \frac{1}{8} \theta \Bigg|_{\frac{\pi}{6}}^{\frac{\pi}{3}} + \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos 2\theta}{8} \, d\theta\)

For the definite integrals:
\(\frac{1}{8} \left(\frac{\pi}{3} - \frac{\pi}{6}\right) = \frac{\pi}{48}\).

\(\int \cos 2\theta \, d\theta = \frac{1}{2} \sin 2\theta\). Evaluate:
\(\frac{1}{16} \left[\sin 2\theta\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{1}{16} (\sin \frac{2\pi}{3} - \sin \frac{\pi}{3}) = 0\) because both angles yield \(\sin \theta = \frac{\sqrt{3}}{2}\).

Thus, \(\iint_G x^2 \, dx \, dy = \frac{\pi}{48}\).

The final calculation is:
\(\frac{200}{\pi} \cdot \frac{\pi}{48} = \frac{200}{48} = \frac{25}{6} \approx 4.1667\).

Therefore, the result is 4.17 (rounded to two decimal places).