**Transcription and Explanation for Educational Use**---**Problem:**1. Suppose \( f(x, y, z) = x^2 + y^2 + z^2 \) and \( W \) is the solid cylinder with height 5 and base radius 2 that is centered about the z-axis with its base at \( z = -2 \). Enter \( \theta \) as theta. (a) As an iterated integral, \[ \iiint\limits_W f \, dV = \int_A^B \int_C^D \int_E^F \left( \boxed{x^2 + y^2 + z^2} \right) \, dz \, dr \, d\theta \] with limits of integration \[ A = \quad \\ B = \quad \\ C = \quad \\ D = \quad \\ E = \quad \\ F = \quad \] (b) Evaluate the integral.**Explanation:**- The problem involves computing a triple integral over a solid cylindrical region. The function is \( f(x, y, z) = x^2 + y^2 + z^2 \).- The cylinder is described as having height 5 and base radius 2. It is aligned with the z-axis, starting from \( z = -2 \).- This problem can be approached by setting up the limits of integration for cylindrical coordinates: - \(\theta\) varies from \(0\) to \(2\pi\). - \(r\) varies from \(0\) to the radius of the cylinder, which is \(2\). - \(z\) varies from \(-2\) to \(3\) (given that the cylinder has a height of 5, starting at \(-2\)).To fill in the integration limits:- \(A = 0\)- \(B = 2\pi\)- \(C = 0\)- \(D = 2\)- \(E = -2\)- \(F = 3\)(b) For evaluating the integral, substitute the function and limits into the integral:\[\int_0^{2\pi} \int_0^2 \int_{-2}^3 (r^2 +

Given, f(x, y, z)=x2+y2+z2, and W is the solid cylinder with a height of 5 and a base radius of 2 that is centered about the z-axis with its base at z=-2.(a). Now, f(x, y, z)=x2+y2+z2, and h=5, r=2, & z=-2. By using cylindrical coordinates: Let, x=rcosθ, y=rsinθ, and z=z. Then, x2+y2=r2, dydx=rdrdθ, & dV=rdzdrdθ, and -2≤z≤3, 0≤r≤2, & 0≤θ≤2π. So, W=r, θ, z) | 0≤r≤2 , 0≤θ≤2π, -2≤z≤3,. Therefore, ∫∫∫WfdV=∫02π∫02∫-23(r2+z2)rdzdrdθ, with the limits of the integration: A= 0 B= 2π C= 0 D= 2 E= -2 F= 3 .(b). ∫∫∫WfdV=∫02π∫02∫-23r2+z2rdzdrdθ, =∫02π∫02∫-23r3+z2rdzdrdθ, =∫02π∫02r3z+13z3r-23drdθ, =∫02π∫023r3+9r--2r3-83rdrdθ, =∫02π∫023r3+9r+2r3+83rdrdθ, =∫02π∫025r3+353rdrdθ, =∫02π54r4+356r202dθ, =∫02π20+703dθ, =1303∫02πdθ, =1303θ02π, =260π3. Hence, ∫∫∫WfdV =260π3.