Volume inside a cone and between z=1 and z=2

[bigevil]

Homework Statement

Write an evaluate a triple integral in spherical coordinates for the volume inside the cone $$z^2 = x^2 + y^2$$ between the planes z=1 and z=2.

The answer is 7π/3

The Attempt at a Solution

Substitute values to work out the limits. From $$z^2 = x^2 + y^2$$, substitute for the spherical coordinates and I get $$\theta = \pi/4$$. Similarly substitute $$1 = x^2 + y^2$$ and $$2^2 = x^2 + y^2$$, and I find that r ranges from $$cosec \theta$$ to $$2 cosec \theta$$.

Then, put together into the triple integral:

$$\int_0^{2\pi = \phi} \int_0^{\pi/4} \int_{cosec\theta}^{2cosec\theta} = r^2 sin\theta dr d\theta d\phi = 2\pi \int_0^{\pi/4} \frac{1}{3}(7 cosec^3 \theta)(sin \theta) d\theta = 2\pi (\frac{7}{3})$$

This answer is off. Appreciate if someone could check the working for me, please!

 

[Unco]

... and I find that r ranges from $$cosec \theta$$ to $$2 cosec \theta$$.

Those cosecs should be secs.