Question

Integrate h(x, y) = yi + xj over the circle of radius 1 centered at the origin traversed counterclockwise.

a)1
b)0
c)pi
d)-1
e)2pi
f)N/A

Answer 1

I think so, but I'm not sure/ This deals with line integrals

Answer 2

thats actually from the problem. i don't really know what to do about the circle with radius 1

Answer 3

the answer is 0, according to my textbook. I don't know why though

Answer 4

You're asked to find
\[\int_C {\bf h}(x,y)\cdot{\bf r}~d{\bf r}\]
where \(C\) is the path of the circle, \({\bf h}(x,y)\) is the given vector field, \(y~\vec{i}+x~\vec{j}\), and \({\bf r}\) is the parameterization for the curve \(C\).

You can go ahead with setting up the line integral, or you can check to see if the vector field is conservative. Conservative vector fields make any line integral over any closed path in the field zero.

Answer 5

so @SithsAndGiggles how do you parametrize r? Because a circle is x^2+y^2=r^2, would the parameterization be u^2+u^2 ?

Answer 6

Not quite, you'll want to use a different parameter for \(x\) and \(y\).

Given a circle \(x^2+y^2=r^2\), you can express it in terms of the Pythagorean identity by setting \(x=r\cos t\) and \(y=r\sin t\), since
\[x^2+y^2=r^2\cos^2t+r^2\sin^2t=r^2\]
with \(t\) ranging from \(0\) to \(2\pi\).

I recommend checking to see if the vector field is conservative first, though setting up the line integral is still good practice.

Answer 7

I'm pretty sure the "vector field" is conservative. So r(u)=\[r^{2}\cos^2 (u)+r ^{2}\sin^2(u)?\]

Answer 8

i'm using u because that's how our book does it; with r(u)

Answer 9

You want \({\bf r}(u)\) to be a vector-valued function, so you should have
\[{\bf r}(u)=\langle \cos u,~\sin u\rangle=\cos u~\vec{i}+\sin u~\vec{j}\\
d{\bf r}(u)=-\sin u~du~\vec{i}+\cos u~du~\vec{j}\]
and using this parameterization means
\[{\bf h}(x(u),y(u))=\langle \cos u,~\sin u\rangle=\cos u~\vec{i}+\sin u~\vec{j}\]
as well.

Since \(0\le u\le2\pi\), you have
\[\int_C{\bf h}\cdot{\bf r}~d{\bf r}=\int_0^{2\pi}(\cos u~\vec{i}+\sin u~\vec{j})\cdot(-\sin u~\vec{i}+\cos u~\vec{j})~du\]

Answer 10

wow that looks hard.

Answer 11

ok so following your method, i got 0.

Answer 12

alright thanks @SithsAndGiggles

Answer 13

You're welcome. The easier way would be to check the partial derivatives of the vector field components. If \({\bf h}(x,y)=M(x,y)\vec{i}+N(x,y)\vec{j}\), then the vector field is conservative if
\[\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\]
which is true in this case, since both partials come out to zero.

Answer 14

alright @SithsAndGiggles . good to know

Answer 15

The only thing I would add is to be mindful of your notation

\[\int_C \mathbf{h}(\mathbf{r})\cdot\,d\mathbf{r} = \int_a^b \mathbf{h}(\mathbf{r}(u))\cdot\mathbf{r}'(u)\,du.\]

Answer 16

thanks @Zarkon. I have that formula in my notes too.