Solid volume using double integrals

Question

I'm trying to calculate the volume of solid bounded by $$z=3\arctan(\frac{y}{x}), z=0, \sqrt{x^2+y^2}=2\arctan(\frac{y}{x})$$ using double integrals, but I dont know where to start.

Answer 1

The volume V can be found in the integral $$\iint_\omega 3 \tan^{-1}\bigg(\frac{y}{x}\bigg )dydx$$

Where $\omega$ is the area bounded by $\sqrt{x^2 + y^2} = 2 \tan^{-1}(\frac{y}{x})$

Let $u = \sqrt{x^2 + y^2}$ and $u$ is between $0$ and $\pi$

$v = \frac{y}{x}$ and $v$ is between $0$ and $\infty$.

Then J : $$J = \bigg(\frac{\partial x}{\partial u}\bigg)\bigg(\frac{\partial y}{\partial v}\bigg ) - \bigg (\frac{\partial x}{\partial v}\bigg ) \bigg(\frac{\partial y}{\partial u}\bigg)$$

$$= \frac{1}{\sqrt{1 + v^2}}\frac{u}{(1 + v^2)^{\frac{3}{2}}} - \frac{-uv}{(1 + v^2)^{\frac{3}{2}}}\frac{v}{\sqrt{1 + v^2}}$$ $$= \frac{u}{1 + v^2}$$

$$V = \int_{0}^{\infty}\int_{0}^{2\tan^{-1}v} 3\tan^{-1}v \frac{u}{1 + v^2} dudv$$

$$= \int_{0}^{\infty}\frac{3}{2} 4(\tan^{-1}v)^{3} \frac{dv}{1 + v^2}$$

Let $v = \tan t$ and $dv = \sec^2 t dt$

$$V = \int_{0}^{\frac{\pi}{2}} 6t^3 dt = \frac{3\pi^4}{32}$$

I have a book with the same question but more general. It asks to find the volume for $$z = c tan^{-1} (\frac{y}{x})$$ And $$\sqrt{x^2 + y^2} = a tan^{-1} (\frac{y}{x})$$

And the answer is $$\frac{\pi^4a^2c}{128}$$ ]2

Answer 2

This the answer for question No 4033 in my book.