Evaluate: ∫02∫23∫0𝑟cos⁡𝜃+𝑟sin⁡𝜃+5(𝑟cos⁡𝜃)𝑟𝑑𝑧𝑑𝑟𝑑𝜃Group of answer choices65𝜋/335𝜋/465𝜋/475𝜋/4 PreviousNext No

The given integral is in cylindrical coordinates. The limits of the integral suggest that we are integrating over a cylindrical volume. The limits for z are from 0 to r, for r are from 2 to 3, and for θ are from 0 to 2π.

The integrand is rcos(θ) + rsin(θ) + 5(rcos(θ)) which simplifies to 6rcos(θ) + rsin(θ).

We can integrate this step by step:

1. First, integrate with respect to z. The integral of a constant with respect to z is just the constant times z. So, we get ∫02∫23[6rcos(θ)z + rsin(θ)z] evaluated from z=0 to z=r drdθ.

2. Substituting the limits for z, we get ∫02∫23[6r^2cos(θ) + r^2sin(θ)] drdθ.

3. Now, integrate with respect to r. The integral of r^2 with respect to r is (1/3)r^3. So, we get ∫02[(1/3)*6r^3cos(θ) + (1/3)r^3sin(θ)] evaluated from r=2 to r=3 dθ.

4. Substituting the limits for r, we get ∫02[(54cos(θ) - 16cos(θ)) + (27sin(θ) - 8sin(θ))] dθ = ∫02[38cos(θ) + 19sin(θ)] dθ.

5. Finally, integrate with respect to θ. The integral of cos(θ) with respect to θ is sin(θ), and the integral of sin(θ) with respect to θ is -cos(θ). So, we get [38sin(θ) - 19cos(θ)] evaluated from θ=0 to θ=2π.

6. Substituting the limits for θ, we get 38sin(2π) - 19cos(2π) - 38sin(0) + 19cos(0) = 0 - 191 - 0 + 191 = 0.

So, the value of the given triple integral is 0.