11. let s be the part of the cylinder ( y^{2}+z^{2}=1 ) that lies between the planes ( x = 0 ) and ( x = 3 )…

Explanation:

Step1: Parameterize the surface

The cylinder (y^{2}+z^{2}=1) can be parameterized as (\vec{r}(x,\theta)=\left\langle x,\cos\theta,\sin\theta\right\rangle), where (0\leq x\leq3) and (0\leq\theta\leq\frac{\pi}{2}) (first - octant).

The partial derivatives are: (\vec{r}{x}=\left\langle1,0,0\right\rangle) (\vec{r}{\theta}=\left\langle0,-\sin\theta,\cos\theta\right\rangle)

The cross - product (\vec{r}{x}\times\vec{r}{\theta}=\left\langle0,-\cos\theta,-\sin\theta\right\rangle)

The magnitude (\left|\vec{r}{x}\times\vec{r}{\theta}\right| = 1)

Step2: Substitute into the surface integral

The integrand (z + 2xy=\sin\theta+2x\cos\theta)

The surface integral (\iint_{S}(z + 2xy)d\sigma=\int_{0}^{3}\int_{0}^{\frac{\pi}{2}}(\sin\theta+2x\cos\theta)d\theta dx)

First, integrate with respect to (\theta): (\int_{0}^{\frac{\pi}{2}}(\sin\theta+2x\cos\theta)d\theta=\left[-\cos\theta + 2x\sin\theta\right]_{0}^{\frac{\pi}{2}}=(0 + 2x)-( - 1+0)=2x + 1)

Step3: Integrate with respect to (x)

(\int_{0}^{3}(2x + 1)dx=\left[x^{2}+x\right]_{0}^{3}=9 + 3=12)

Answer:

(12)