Question

The volume integral of the function \( f(r, \theta, \phi) = r^2 \cos\theta \) over the region \( 0 \leq r \leq 2, 0 \leq \theta \leq \frac{\pi}{3} \) and \( 0 \leq \phi \leq 2\pi \) is:

Hint:

When solving volume integrals in spherical coordinates, make sure to include the appropriate \( r^2 \sin\theta \) factor and adjust limits accordingly.

Answer 1

Step 1: Set up the integral.
We are given the function \( f(r, \theta, \phi) = r^2 \cos\theta \) and the region of integration \( 0 \leq r \leq 2, 0 \leq \theta \leq \frac{\pi}{3}, 0 \leq \phi \leq 2\pi \). The volume integral is: \[ I = \int_0^{2\pi} \int_0^{\frac{\pi}{3}} \int_0^2 r^2 \cos\theta \, r^2 \sin\theta \, dr \, d\theta \, d\phi \] Step 2: Perform the integration over \( r \).
The first part of the integral involves the \( r \) terms: \[ \int_0^2 r^4 \, dr = \frac{r^5}{5} \Bigg|_0^2 = \frac{32}{5} \] Step 3: Perform the integration over \( \theta \).
Now integrate with respect to \( \theta \): \[ \int_0^{\frac{\pi}{3}} \cos\theta \sin\theta \, d\theta = \frac{1}{2} \int_0^{\frac{\pi}{3}} \sin(2\theta) \, d\theta \] This results in: \[ \frac{1}{2} \left[ -\frac{\cos(2\theta)}{2} \right]_0^{\frac{\pi}{3}} = \frac{1}{2} \left( -\frac{\cos\left(\frac{2\pi}{3}\right)}{2} + \frac{1}{2} \right) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \] Step 4: Perform the integration over \( \phi \).
Now integrate over \( \phi \): \[ \int_0^{2\pi} d\phi = 2\pi \] Step 5: Final Calculation.
Now, combining all parts: \[ I = \frac{32}{5} \times \frac{1}{4} \times 2\pi = \frac{16\pi}{5} \] Step 6: Conclusion.
Thus, the value of the integral is \( \frac{16\pi}{5} \).