Compute Curl Integral of Vector Field and Ellipse

Question

Consider a vector field : ${\bf F}=P(x){\bf i}+ Q(y){\bf j}$, for some functions : $P(x), Q(y)$ which have continuous partial derivatives everywhere.

Let: $C$ stand for the ellipse : $\{ 4x^2+9y^2=1\}$. Then : $\int_C {\bf F}\cdot d{\bf r}$ equals:

I can't determine an answer with this data but the options are: $P(\frac{1}{2}) - Q(\frac{1}{2}), 0, P(\frac{1}{2}),P(\frac{1}{2}) + Q(\frac{1}{2})$

Which is the right answer?

score 1 · Answer 1 · answered Apr 17 '20 at 06:48

1

Notice that

$\nabla \times \mathbf F = \dfrac{\partial Q(y)}{\partial x} - \dfrac{\partial P(x)}{\partial y} = 0; \tag 1$

next recall that, via Stokes theorem,

$\displaystyle \int_\xi \nabla \times \mathbf F dA = \int_{\partial \xi} \mathbf F \cdot d \mathbf r = \int_C \mathbf F \cdot d \mathbf r, \tag 2$

since

$\partial \xi = C; \tag 3$

here $\xi$ is the region bounded by the ellipse $C$, and $dA$ is the area element on $\Bbb R^2$. Together (1) and (2) yield

$\displaystyle \int_C \mathbf F \cdot d \mathbf r = 0, \tag 4$

which is thus the correct option.

answered Apr 17 '20 at 06:48

Robert Lewis

71,180

1

How do you obtain that the field is irrotational when we don't know what $P$ and $Q$ are? – Allawonder Apr 17 '20 at 06:52
@Allawonder; well, $P(x)$ usually means $P$ is a function of $x$ alone, not of $y$, so $\partial P(x) / \partial y = $ and so forth. Cheers! – Robert Lewis Apr 17 '20 at 06:54
1

Oh, of course! LOL. – Allawonder Apr 17 '20 at 07:04

score 0 · Answer 2 · answered Apr 17 '20 at 06:37

0

Hint. You're supposed to use Green's theorem, so that your integral is equal to $$\iint_{\mathcal E} (Q_x-P_y)\mathrm dx \mathrm dy,$$ with $\mathcal E$ being the region determined by the ellipse. Can you continue now?

answered Apr 17 '20 at 06:37

Allawonder

13,327

Compute Curl Integral of Vector Field and Ellipse

2 Answers2