Let's break the problem into **two clear steps**, as requested: --- ## **Step 1: Setting up the Integral in Cylindrical Coordinates** ### **Given Region** The solid cubic region: \[ E = [,1] \times [,1] \times [,1] \] in \(\mathbb{R}^3\). ### **Cylindrical Coordinates** Recall the conversions: - \(x = r\cos\theta\) - \(y = r\sin\theta\) - \(z = z\) - \(dV = r\, dr\, d\theta\, dz\) #### **Describing the Cube in Cylindrical Coordinates** - \(z\) runs from \(\) to \(1\). - \(x\) and \(y\) are both from \(\) to \(1\), which in the \(xy\)-plane forms a square, **not a full circle**—so the limits for \(r\) and \(\theta\) are nonstandard: - \(\theta\) runs from \(\) to \(\frac{\pi}{2}\) (since both \(x \geq \) and \(y \geq \)). - For each \(\theta\), \(r\) runs from \(\) to the minimum of \(x=1\) and \(y=1\): So, \(r \leq \min \left( \frac{1}{\cos\theta}, \frac{1}{\sin\theta} \right)\). But since \(\cos\theta, \sin\theta > \) in \([, \frac{\pi}{2}]\), the upper bound is \(r = \min(1/\cos\theta, 1/\sin\theta)\). However, in the unit square, the boundary for \(r\) is where \(x = 1\) or \(y = 1\): - For a given \(\theta\), \(r\) goes from \(\) to \(\min(1/\cos\theta, 1/\sin\theta)\). #### **Cylindrical Integral Setup** \[ \iiint_E f \, dV = \int_{z=}^1 \int_{\theta=}^{\frac{\pi}{2}} \int_{r=}^{\min(1/\cos\theta,1/\sin\theta)} f(r\cos\theta, r\sin\theta, z) \, r \, dr \, d\theta \, dz \] **Explanation:** - The bounds for \(z\) reflect the cube's height. - The bounds for \(\theta\) reflect the positive quadrant. - The bounds for \(r\) are found by the edges of the square in the \(xy\)-plane. --- ## **Step 2: Setting up the Integral in Spherical Coordinates** ### **Spherical Coordinates** Recall the conversions: - \(x = \rho \sin\phi \cos\theta\) - \(y = \rho \sin\phi \sin\theta\) - \(z = \rho \cos\phi\) - \(dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta\) #### **Describing the Cube in Spherical Coordinates** The cube is not aligned with spheres, so the limits are awkward: - \(\theta\) runs from \(\) to \(\frac{\pi}{2}\) (first quadrant). - \(\phi\) runs from \(\) to \(\arccos() = \frac{\pi}{2}\) (since \(z \geq \)). - For each \((\theta, \phi)\), \(\rho\) runs from \(\) up to the minimum value so that \(x, y, z \leq 1\): \[ \begin{align*} x &= \rho \sin\phi \cos\theta \leq 1 \implies \rho \leq \frac{1}{\sin\phi \cos\theta} \\ y &= \rho \sin\phi \sin\theta \leq 1 \implies \rho \leq \frac{1}{\sin\phi \sin\theta} \\ z &= \rho \cos\phi \leq 1 \implies \rho \leq \frac{1}{\cos\phi} \end{align*} \] Thus, \[ \rho \leq \min\left( \frac{1}{\sin\phi \cos\theta}, \frac{1}{\sin\phi \sin\theta}, \frac{1}{\cos\phi} \right) \] #### **Spherical Integral Setup** \[ \iiint_E f \, dV = \int_{\theta=}^{\frac{\pi}{2}} \int_{\phi=}^{\frac{\pi}{2}} \int_{\rho=}^{\rho_{\text{max}}(\theta, \phi)} f(\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi) \, \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \] Where \[ \rho_{\text{max}}(\theta, \phi)=\min\left( \frac{1}{\sin\phi \cos\theta}, \frac{1}{\sin\phi \sin\theta}, \frac{1}{\cos\phi} \right) \] **Explanation:** - The bounds for \(\theta\) and \(\phi\) cover the positive axes. - The bounds for \(\rho\) ensure that \(x, y, z\) stay within the cube. --- ## **Final Answers** **(a) Cylindrical Coordinates:** \[ \boxed{ \iiint_E f \, dV = \int_{z=}^1 \int_{\theta=}^{\frac{\pi}{2}} \int_{r=}^{\min(1/\cos\theta,1/\sin\theta)} f(r\cos\theta, r\sin\theta, z) \, r \, dr \, d\theta \, dz } \] **(b) Spherical Coordinates:** \[ \boxed{ \iiint_E f \, dV = \int_{\theta=}^{\frac{\pi}{2}} \int_{\phi=}^{\frac{\pi}{2}} \int_{\rho=}^{\min\left( \frac{1}{\sin\phi \cos\theta}, \frac{1}{\sin\phi \sin\theta}, \frac{1}{\cos\phi} \right)} f(\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi) \, \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta } \] --- **Each step explains how the region is described in the respective coordinates, and how the bounds are found.**