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Chapter 12: Problem 4

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

Expert verified
The value of the given iterated integral is 0.

Step by step solution

01

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:\[\frac{1}{3} \cos^{3} \theta \sin \phi \cos \phi\]
02

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:\[\frac{1}{12} \sin \phi \cos \phi\]
03

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:\[-\frac{1}{48} + \frac{1}{48} = 0\]

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Most popular questions from this chapter

In Exercises \(31-36,\) use an iterated integral to find the area of the region bounded by the graphs of the equations. $$ \sqrt{x}+\sqrt{y}=2, \quad x=0, \quad y=0 $$

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