To evaluate the triple iterated integral

$\int_0^{\pi/2} \int_0^{\pi/4} \int_0^{\cos \theta} \rho^2 \sin \varphi \cos \varphi \, d\rho \, d\theta \, d\varphi,$

let's proceed by integrating in the order given: first with respect to $\rho$, then $\theta$, and finally $\varphi$.

Step 1: Integrate with respect to $\rho$

The inner integral is:

$\int_0^{\cos \theta} \rho^2 \sin \varphi \cos \varphi \, d\rho.$

Since $\sin \varphi \cos \varphi$ is treated as a constant with respect to $\rho$, this becomes:

$\sin \varphi \cos \varphi \int_0^{\cos \theta} \rho^2 \, d\rho.$

Now, integrate $\rho^2$ with respect to $\rho$:

$\int_0^{\cos \theta} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_0^{\cos \theta} = \frac{(\cos \theta)^3}{3}.$

Thus, the inner integral simplifies to:

$\int_0^{\cos \theta} \rho^2 \sin \varphi \cos \varphi \, d\rho = \frac{(\cos \theta)^3}{3} \sin \varphi \cos \varphi.$

Step 2: Integrate with respect to $\theta$

Now we have:

$\int_0^{\pi/2} \int_0^{\pi/4} \frac{(\cos \theta)^3}{3} \sin \varphi \cos \varphi \, d\theta \, d\varphi.$

This separates as:

$\frac{1}{3} \int_0^{\pi/2} \int_0^{\pi/4} (\cos \theta)^3 \sin \varphi \cos \varphi \, d\theta \, d\varphi.$

The integral with respect to $\theta$ is:

$\int_0^{\pi/4} (\cos \theta)^3 \, d\theta.$

To evaluate this, use the reduction formula for powers of cosine, or recognize it as a standard integral. The result of this integral is:

$\int_0^{\pi/4} (\cos \theta)^3 \, d\theta = \frac{3 \pi}{32}.$

So the expression becomes:

$\frac{1}{3} \cdot \frac{3 \pi}{32} \int_0^{\pi/2} \sin \varphi \cos \varphi \, d\varphi = \frac{\pi}{32} \int_0^{\pi/2} \sin \varphi \cos \varphi \, d\varphi.$

Step 3: Integrate with respect to $\varphi$

Now we have:

$\frac{\pi}{32} \int_0^{\pi/2} \sin \varphi \cos \varphi \, d\varphi.$

Using the identity $\sin \varphi \cos \varphi = \frac{1}{2} \sin(2\varphi)$, this becomes:

$\frac{\pi}{32} \cdot \frac{1}{2} \int_0^{\pi/2} \sin(2\varphi) \, d\varphi = \frac{\pi}{64} \int_0^{\pi/2} \sin(2\varphi) \, d\varphi.$

Now, integrate $\sin(2\varphi)$:

$\int_0^{\pi/2} \sin(2\varphi) \, d\varphi = \left[ -\frac{1}{2} \cos(2\varphi) \right]_0^{\pi/2} = \frac{1}{2}.$

Thus, the entire expression becomes:

$\frac{\pi}{64} \cdot \frac{1}{2} = \frac{\pi}{128}.$

Final Answer

$\int_0^{\pi/2} \int_0^{\pi/4} \int_0^{\cos \theta} \rho^2 \sin \varphi \cos \varphi \, d\rho \, d\theta \, d\varphi = \frac{\pi}{128}.$

Would you like more details on any of the steps?

1. What is the reduction formula for powers of cosine?
2. How can trigonometric identities simplify integration?
3. Why do we change integration order in triple integrals sometimes?
4. Can integration by parts be used here, and if so, how?
5. How do we evaluate trigonometric integrals more efficiently?

Tip: When dealing with trigonometric integrals, always look for identities that can simplify products into single terms.