Question

The value of \[ \int_0^{\frac{\pi}{2}} \int_0^{\cos \theta} r \sin \theta \, dr \, d\theta \] is

• A. 0
• B. \( \frac{1}{6} \)
• C. \( \frac{4}{3} \)
• D. \( \pi \)

Hint:

For double integrals, first integrate over the inner variable and then over the outer variable. Use substitution if necessary to simplify the integration.

Answer 1

The Correct Option is B

We are asked to evaluate the double integral: \[ I = \int_0^{\frac{\pi}{2}} \int_0^{\cos \theta} r \sin \theta \, dr \, d\theta. \] Step 1: Integrate with respect to \( r \) First, perform the integration over \( r \). The inner integral is: \[ \int_0^{\cos \theta} r \, dr = \frac{r^2}{2} \Big|_0^{\cos \theta} = \frac{\cos^2 \theta}{2}. \] Thus, the double integral becomes: \[ I = \int_0^{\frac{\pi}{2}} \frac{\cos^2 \theta}{2} \sin \theta \, d\theta. \] Step 2: Simplify and integrate with respect to \( \theta \) We now need to integrate: \[ I = \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos^2 \theta \sin \theta \, d\theta. \] We can use the substitution \( u = \cos \theta \), so \( du = -\sin \theta \, d\theta \). The limits change as follows: - When \( \theta = 0 \), \( u = 1 \), - When \( \theta = \frac{\pi}{2} \), \( u = 0 \). The integral becomes: \[ I = \frac{1}{2} \int_1^0 u^2 (-du) = \frac{1}{2} \int_0^1 u^2 \, du. \] Step 3: Evaluate the integral Now, integrate: \[ \int_0^1 u^2 \, du = \frac{u^3}{3} \Big|_0^1 = \frac{1}{3}. \] Thus, the value of the double integral is: \[ I = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}. \] Therefore, the correct answer is Option (B).
Final Answer: (B) \( \frac{1}{6} \)