Question

Evaluate the iterated integral. $$ \int_{0}^{\pi / 4} \int_{0}^{\pi / 4} \int_{0}^{\cos \theta} \rho^{2} \sin \phi \cos \phi d \rho d \theta d \phi $$

Short answer

The value of the given iterated integral is 0.

Step-by-step solution

Evaluate the innermost integral

Calculate the innermost integral with respect to \( \rho \):\[\int_{0}^{\cos \theta} \rho^{2} \sin \phi \cos \phi d \rho \] This is pretty straightforward since we just need to apply the power rule of integration. The result is:\[\frac{1}{3} \rho^{3} \sin \phi \cos \phi \Biggr|_0^{\cos \theta}\] This simplifies to:\[\frac{1}{3} \cos^{3} \theta \sin \phi \cos \phi\]

Evaluate the second integral

Now consider the second integral with respect to \( \theta \):\[\int_{0}^{\pi / 4} \frac{1}{3} \cos^{3} \theta \sin \phi \cos \phi d \theta\] This is also a direct application of the power rule of integration. The result is:\[\frac{1}{12} \cos^{4} \theta \sin \phi \cos \phi \Biggr|_0^{\frac{\pi}{4}}\] This simplifies to:\[\frac{1}{12} \sin \phi \cos \phi\]

Evaluate the outermost integral

Finally, evaluate the outermost integral with respect to \( \phi \):\[\int_{0}^{\pi / 4} \frac{1}{12} \sin \phi \cos \phi d \phi \] We can use the double angle formula \( \sin 2x = 2 \sin x \cos x \) to simplify the integrand to \( \frac{1}{24} \sin 2 \phi \). Now it's another direct application of the power rule of integration. The result is:\[-\frac{1}{48} \cos 2 \phi \Biggr|_0^{\frac{\pi}{4}}\] This simplifies to:\[-\frac{1}{48} + \frac{1}{48} = 0\]