Multivariable Calculus Question of the Day

1. $\int_0^3 \int_{-2}^2 yz \sqrt{1 + \left(\frac{-x}{\sqrt{4-x^2}}\right)^2} \, dx \, dy$
2. $\int_0^3 \int_0^\pi 2y\sin\theta \cdot 2 \, d\theta \, dy$ (correct answer)
3. $\int_0^3 \int_{-2}^2 yz \, dx \, dy$ where $z = \sqrt{4-x^2}$
4. $\int_0^\pi \int_0^3 2y\sin\theta \, dy \, d\theta$

Explanation: Using cylindrical coordinates with $x = 2\cos\theta$, $z = 2\sin\theta$ for $0 \leq \theta \leq \pi$ and $0 \leq y \leq 3$. The parameterization is $\mathbf{r}(\theta,y) = (2\cos\theta, y, 2\sin\theta)$. We get $\mathbf{r}_\theta = (-2\sin\theta, 0, 2\cos\theta)$ and $\mathbf{r}_y = (0, 1, 0)$. The magnitude $|\mathbf{r}_\theta \times \mathbf{r}_y| = 2$. On the surface, $g = yz = y \cdot 2\sin\theta$. The integral becomes $\int_0^3 \int_0^\pi 2y\sin\theta \cdot 2 \, d\theta \, dy$. Choice A uses the wrong surface element calculation. Choice C omits the surface element entirely. Choice D has the correct integrand but wrong bounds and missing factor of 2.