Question

Consider the surface \[ S = \{(x,y,xy) \in \mathbb{R}^3 : x^2 + y^2 \le 1\}. \] Let \(\vec{F} = y\hat{i} + x\hat{j} + \hat{k}\). If \(\hat{n}\) is the continuous unit normal field to the surface \(S\) with positive \(z\)-component, then \[ \iint_S \vec{F} \cdot \hat{n} \, dS \] equals

• A. \(\frac{\pi}{4}\)
• B. \(\frac{\pi}{2}\)
• C. \(\pi\)
• D. \(2\pi\)

Hint:

For surfaces of the form \(z=f(x,y)\), use the formula \(\iint_S \vec{F}\cdot\hat{n}\,dS = \iint_R \vec{F}\cdot(-f_x,-f_y,1)\,dx\,dy\).

Answer 1

The Correct Option is B

Step 1: Parameterize the surface.
Let \(z = xy\). Then \[ \vec{r}(x,y) = x\hat{i} + y\hat{j} + xy\hat{k}. \] Compute \[ \vec{r}_x = \hat{i} + y\hat{k}, \quad \vec{r}_y = \hat{j} + x\hat{k}. \] The normal vector is \[ \vec{r}_x \times \vec{r}_y = (\hat{i} + y\hat{k}) \times (\hat{j} + x\hat{k}) = (y^2 - x^2)\hat{k} - y\hat{i} - x\hat{j}. \]
Step 2: Compute \(\vec{F}\cdot\hat{n}\).
Using the positive \(z\)-component convention, \[ \vec{F}\cdot\hat{n} = (y, x, 1) \cdot (-y, -x, 1) = -x^2 - y^2 + 1. \]
Step 3: Evaluate the integral.
\[ \iint_S \vec{F}\cdot\hat{n} \, dS = \iint_{x^2+y^2\le1} (1 - x^2 - y^2) \, dx\,dy. \] In polar coordinates, \[ \int_0^{2\pi}\int_0^1 (1 - r^2)r\,dr\,d\theta = 2\pi\int_0^1 (r - r^3)dr = 2\pi\left(\frac{1}{2} - \frac{1}{4}\right) = \frac{\pi}{2}. \]
Step 4: Conclusion.
Hence, the value of the surface integral is \(\boxed{\frac{\pi}{2}}\).