Question

The value of the triple integral \( \iiint_V (x^2y + 1) \, dx\,dy\,dz \), where \( V \) is the region given by \( x^2 + y^2 \le 1, \, 0 \le z \le 2, \) is

• A. \( \pi \)
• B. \( 2\pi \)
• C. \( 3\pi \)
• D. \( 4\pi \)

Hint:

Always check for terms that vanish over a full revolution when integrating trigonometric functions in cylindrical coordinates.

Answer 1

The Correct Option is B

Step 1: Express the region in cylindrical coordinates. 
\[ x = r\cos\theta, \, y = r\sin\theta, \, z = z, 0 \le r \le 1, \, 0 \le \theta \le 2\pi, \, 0 \le z \le 2. \] \[ x^2y + 1 = r^2\cos^2\theta (r\sin\theta) + 1 = r^3\cos^2\theta\sin\theta + 1. \]

Step 2: Write the integral. 
\[ \iiint_V (x^2y + 1) \, dV = \int_0^{2\pi} \int_0^1 \int_0^2 (r^3\cos^2\theta\sin\theta + 1)r \, dz\,dr\,d\theta. \]

Step 3: Integrate with respect to \( z \). 
\[ = \int_0^{2\pi} \int_0^1 [2r^4\cos^2\theta\sin\theta + 2r] \, dr\,d\theta. \] The first term integrates to zero because \( \int_0^{2\pi}\cos^2\theta\sin\theta \, d\theta = 0. \) \[ \Rightarrow \int_0^{2\pi}\int_0^1 2r\, dr\, d\theta = 2\pi. \]

Final Answer: \( 2\pi. \)