91Ó°ÊÓ

$$F(x, y,z) = yi + x j + k$$, where $$S$$ is the paraboloid with equation $$z = 4x^{2} + y^{2} + 1$$ that lies above the annulus determined by $$1 ≤ x^{2} + y^{2} ≤ 4$$ in the xy-plane.

Partially differentiating $$z$$ with respect to $$x$$, we get

$$\frac{\partial z}{\partial x}=8x$$

Partially differentiating $$z$$ with respect to $$y$$, we get

$$\frac{\partial z}{\partial y}=2y$$

So, $$dS= \sqrt{64x^{2}+4y^{2}+1}dA$$

Also, we get the unit vector $$n$$ as,

$$n=\frac{\langle -8x, -2y, 1 \rangle}{\sqrt{64x^{2}+4y^{2}+1}}$$

Now, evaluating the integral, we get

$$\int \int F\cdot n dS=\int \int \frac{1}{\sqrt{64x^{2}+4y^{2}+1}}(-8xy-2xy+1)$$

$$\implies \int \int F\cdot n dS=\int \int (-10xy+1)dxdy$$

Solving further, we get

$$\int \int (-10xy+1)dxdy =\int_{0}^{2\pi}\int_{1}^{4}(-10r^{2}sin \theta cos \theta)rdrd\theta$$

$$\implies \int \int (-10xy+1)dxdy=3 \pi$$