# Can someone please help me verify if this is the correct way to set up an integral in spherical coordinates?

Let $$W$$ be the ice cream cone

$$W= \Big\{(x,y,z) | \sqrt{3x^2+3y^2} \leq z \leq \sqrt{1-x^2-y^2}\Big\}$$ in spherical coordinates.

$$\int_{0}^{2\pi} \int_{0}^{\pi/6}\int_{0}^{1} \rho^2 \sin\phi d\rho d\phi d\theta$$

$$\int_{0}^{2\pi} \int_{0}^{\pi/6}\int_{1}^{0} \rho^2 \sin\phi d\rho d\phi d\theta$$

Since the region looks like this: And the radius goes from 1 to 0?

I can show my work for the remainder of the problem, but essentially I am just stuck on this one part for the bounds of $$\rho$$

• Is the integral you're trying to evaluate $$\iiint_W\sqrt{x^2+y^2+z^2}\,dxdydz\ ?$$ That's what your first integral in polar coordinates is equal to. Your second integral would evaluate to its negative. Jun 7, 2022 at 6:13

Using $$\rho$$ for radius, $$\phi$$ for latitude and $$\theta$$ for longitude:
$$x = \rho \cos\phi \cos\theta$$ $$y = \rho \cos\phi \sin\theta$$ $$z = \rho \sin\phi$$ $$\rho\cos\phi \sqrt{3} \leq \rho\sin\phi\leq\sqrt{1-\rho^2\cos^2\phi} \Longleftrightarrow$$ $$\Longleftrightarrow \tan \phi \geq \sqrt{3} \wedge \rho \leq 1 \Longleftrightarrow \frac{\pi}{3} \leq \phi \leq \frac{\pi}{2} \wedge \rho \leq 1$$ And since in spehrical coordinates $$dV = \rho^2 \cos\phi \;d\rho \;d\phi\;d\theta$$ we get:
$$V = \int_0^{2\pi} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \int_0^1 \rho^2 \cos\phi \;d\rho \;d\phi\;d\theta \;\;=\pi\frac{2-\sqrt{3}}{3}$$