<s>[INST] Finding error in evaluation of triple integral The following problem is from lecture 26 of MIT OCW's multivariable calculus:

Evaluate $$\int_0^{2\pi} \int_{0}^{\frac{\pi}{4}} \int_{\frac{\sec \phi}{\sqrt{2}}}^1 \rho^2 \sin \phi\; d\rho\; d\phi\; d\theta$$ 

For some reason, I keep getting a different answer to the lecturer. Furthermore, I get a negative answer although the integral is supposed to describe the volume of a segment of a sphere. I have been unable to figure out what I am doing wrong. This is how I have done the problem:
Inner integral:
$$\int_{\frac{\sec \phi}{\sqrt{2}}}^1 \rho^2 \sin \phi\; d\rho=\sin \phi \cdot [\frac{\rho^3}{3}]_{\frac{\sec \phi}{\sqrt{2}}}^1=\frac{\sin \phi}{3}-\frac{\sin \phi \cdot \sec^3 \phi}{6\sqrt{2}}$$
Middle Integral:
$$\int_{0}^{\frac{\pi}{4}}\frac{\sin \phi}{3}-\frac{\sin \phi \cdot \sec^3 \phi}{6\sqrt{2}}\;d\phi=[-\frac{\cos \phi}{3}-\frac{\tan^2 \phi}{12\sqrt{2}}]_0^{\frac{\pi}{4}}=-\frac{\sqrt{2}}{6}-\frac{1}{12\sqrt{2}}$$
Outer Integral:
$$\int_0^{2\pi}-\frac{\sqrt{2}}{6}-\frac{1}{12\sqrt{2}} \; d\theta=-\frac{2\pi\sqrt{2}}{6}-\frac{2\pi}{12\sqrt{2}}$$
Which simplifies to $$-\frac{\pi \sqrt{2}}{3}-\frac{\pi}{6\sqrt{2}}=-(\frac{12\pi+3\pi}{18\sqrt{2}})=-\frac{5\pi}{6\sqrt{2}}$$
The answer that the lecturer gets is $$\frac{2\pi}{3}-\frac{5\pi}{6\sqrt{2}}$$
Could anyone tell me what I have done wrong? Any help would be greatly appreciated. 
 [/INST] In the middle integral: $\cos 0 = 1$. This gives you a $+ \frac{1}{3}$, which integrated in the next step adds $\frac{2\pi}{3}$, as desired.
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