Given:
$$D = \left\{(x,y) | 1 \leq x^2 +y^2 \leq 100, ~~ \frac{x \sqrt{3}}{3} \leq y \leq x \sqrt{3} \right\}$$
Calculate the Volume under the surface $$z = x^2 + y^2$$
And above $D$.
My try:
$$I = \iint_D x^2 + y^2 \,dx\,dy$$
We can write $D$ as:
$$D = \left\{(x,y) | 1 \leq x^2 +y^2 \leq 100, ~~ \frac{\sqrt{3}}{3} \leq \frac{y}{x} \leq \sqrt{3} \right\}$$
And change to $u,v$ as the following:
$$\left\{\begin{matrix} u = x^2 + y^2 \\ v = \frac{y}{x} \end{matrix}\right.$$
So the set $D$ would be:
$$D = \left\{(u,v) | 1 \leq u \leq 100, ~~ \frac{1}{\sqrt{3}} \leq v \leq \sqrt{3} \right\}$$
Now, because we changed the variables, we need to calculate the Jacobian:
$$ J = \frac{ D(x,y)}{D(u,v)} = \frac{1}{\frac{D(u,v)}{D(x,y)}} = \frac{1}{\begin{vmatrix} u_x & u_y\\ v_x & v_y \end{vmatrix}} = \frac{1}{\begin{vmatrix} 2x & 2y\\ -\frac{y}{x^2} & \frac1x \end{vmatrix}} = \frac{1}{2+ 2 (\frac{y}{x})^2}$$
$$J = \frac{1}{2(1+v^2)}$$
And we can calculate the integral as so:
$$\iint_{D_{uv}} u \cdot \frac{1}{2} \cdot \frac{1}{1+v^2} \,du\,dv = \frac{1}{2} \int_1^{100} ( \int_{ \frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{u}{1+v^2} du)dv = \frac{1}{2} \int_1^{100} \frac{1}{1+v^2} ( \frac{ (\sqrt{3})^2 - (\frac{1}{ \sqrt{3}})^2}{2})dv = \frac23 \int_1^{100} \frac{1}{1+v^2} dv = \frac23 ( \arctan{v} |_1^{100}) \approx 0.5169 \dots$$
Some people I talked with got an answer above $10,000$ and my answer is not close at all! So I ask for your help, if you can review my work, I would be so thankful! Thanks for helping!
In polar coordinates, the region $D$ is equivalent to $$D = \{(r, \theta) : (1 \le r \le 10) \cap (\tfrac{\pi}{6} \le \theta \le \tfrac{\pi}{3})\}.$$ The proof is straightforward after noting that $r^2 = x^2 + y^2$ and $\tan \theta = \frac{y}{x}$. Consequently, the desired volume is easily expressed as an integral in cylindrical coordinates: $$V = \int_{r=1}^{10} \int_{\theta = \pi/6}^{\pi/3} r^3 \, d\theta \, dr.$$