91Ó°ÊÓ

The innermost integral is with respect to \( z \), within the limits from \( 0 \) to \( 1 \). Since the integrand is \( r \sin\theta \), the integral with respect to \( z \) is calculated as: \[ \int_{0}^{1} r \sin\theta \, dz = r \sin\theta [z]_{0}^{1} = r \sin\theta \] Now the integral changes to, \[ \int_{-rac{\pi}{4}}^{\frac{\pi}{4}} \int_{0}^{1-2 \cos^{2} \theta} r \sin\theta \, dr \, d\theta \]

Now we evaluate the integral with respect to \( r \), within the limits from \( 0 \) to \( 1-2 \cos^{2} \theta \). \[ \int_{0}^{1-2 \cos^{2} \theta} r \sin\theta \, dr = \sin\theta \left[ \frac{1}{2} r^2 \right]_{0}^{1-2 \cos^{2} \theta} \]\[ = \frac{1}{2} \sin\theta \left( (1-2 \cos^{2} \theta)^2 - 0^2 \right) \]After squaring the expression and maintaining the \( \sin\theta \) term, we have:\[ = \frac{1}{2} \sin\theta (1 - 4 \cos^{2}\theta + 4 \cos^{4}\theta) \] The integral expression now is, \[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2} \sin\theta (1 - 4 \cos^{2}\theta + 4 \cos^{4}\theta) \, d\theta \]

Now evaluate the integral with respect to \( \theta \). Split the expression and integrate each part separately: \[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \frac{1}{2} \sin\theta - 2 \sin\theta \cos^{2}\theta + 2 \sin\theta \cos^{4}\theta \right) \, d\theta \]Since the integrals of symmetric expressions around the origin with sine functions result in zero due to the sine's odd symmetry, we find:1. \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2} \sin\theta \, d\theta = 0 \) 2. Other terms \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} -2 \sin\theta \cos^{2}\theta \, d\theta + \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 2 \sin\theta \cos^{4}\theta \, d\theta \) also result in zero due to similar symmetry considerations.Thus, the iterated integral evaluates to zero.