Volume of a sphere above a plane

Question

I'm trying to find the volume of a region contained within a sphere centered at the origin of radius 2, and above the plane z=1. I made the computation below and just wanted another pair of eyes to check my bounds and integration. Thanks! $$\int_0^{2\pi}\int_0^{\pi/3}\int_{\sec\phi}^2\rho^2\sin\phi\ d\rho \ d\phi\ d\theta$$ $$\frac13\int_0^{2\pi}\int_0^{\pi/3}8\sin\phi-\sec^3\phi\sin\phi\ d\phi\ d\theta$$ $$\frac13\int_0^{2\pi}\int_0^{\pi/3}8\sin\phi-\sec^2\phi\tan\phi\ d\phi\ d\theta$$ $$\frac56\int_0^{2\pi}\ d\theta$$ $$\frac{5\pi}3$$

Answer 1

Yes the set up is correct and also the calculation indeed

$$\int_0^{2\pi}\,d\theta\int_0^{\pi/3}\,d\phi \int_{\sec\phi}^2\rho^2\sin\phi\ d\rho =2\pi \int_0^{\pi/3}\sin \phi[\rho^3/3]_{\sec\phi}^2\,d\phi=\\=\frac{2\pi}3 \int_0^{\pi/3} 8\sin \phi-\frac{\sin \phi}{\cos^3 \phi}\,d\phi =\frac{2\pi}3\left[-8\cos \phi-\frac12\sec^2 \phi\right]_0^{\pi/3}=\frac{2\pi}3\left(-4-2 +8+\frac12\right)=\frac{2\pi}3\left(\frac52\right)=\frac{5\pi}3$$