The cone has the formula: $x^2 + y^2 = z^2 , 0≤z≤2$
So I used the cylindrical coordinates to get the following answer:
$$\int_0^{2\pi}\int_0^2\int_0^2 dz\,rdr\,d\theta = 8\pi$$
In the solution of the doctor, he used spherical coordinates as follows:
$$\int_0^{2\pi}\int_0^{\pi/4}\int_0^{2\sec\Phi}\rho^2\sin\Phi \,d\rho \,d\Phi \,d\theta=\frac{8\pi}{3}$$
- Why is my answer wrong?
- Why, in doctor's solution, $\rho=2\sec\Phi$ in the upper limit of the integration?
Best Answer
Your answer gives you the volume of a cylinder with height $2$.
See the picture:
$$\frac{2}{\rho}=\cos \phi\implies \rho=\frac{2}{\cos\phi}=2\sec\phi$$