Help setting up two line integrals

**NickE** · July 28th 2012, 07:02 AM

How do I set up these line integrals?

First problem:

Line integral of y^3dx + x^2dy where C is the arc of the parabola x=1-y^2 from (0,-1) to (0,1).

Second problem:

Line integral of ydx + (x+y^2)dy where C is the ellipse 4x^2 + 9y^2 = 36, with counterclockwise rotation

**Prove It** · July 28th 2012, 07:39 AM

Originally Posted by NickE

How do I set up these line integrals?

First problem:

Line integral of y^3dx + x^2dy where C is the arc of the parabola x=1-y^2 from (0,-1) to (0,1).

Second problem:

Line integral of ydx + (x+y^2)dy where C is the ellipse 4x^2 + 9y^2 = 36, with counterclockwise rotation

Have you used Green's Theorem before?

$\displaystyle \begin{align*} \oint_C{\left(L\,dx + M\,dy\right)} &= \int{\int_D{\left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\,dx}\,dy} \end{align*}$

This will help you with the second problem because it's a closed contour.

**NickE** · July 28th 2012, 07:46 AM

Thanks this helps a lot. I knew there was a way to convert a line integral in that form to a double integral but could not find it in my notes.

Edit:
I still need help. I've found my notes on green's theorem but there are no examples on how to apply it to ellipse. I tried turning problem 2 it into a double integral and came up with the double integral of 0. What am I doing wrong. I know I'm probably overthinking this.

**HallsofIvy** · July 28th 2012, 09:35 AM

Originally Posted by NickE

How do I set up these line integrals?

First problem:

Line integral of y^3dx + x^2dy where C is the arc of the parabola x=1-y^2 from (0,-1) to (0,1).

Have you not had any instruction in this? Since x= 1- y^2, dx= -2ydy so y^3dx+ x^2dy= y^3(-2ydy)+ (1- y^2)^2dy= -2y^4dy+ (1- 2y^2+ y^4)dy= (-y^4- 2y^2+ 1)dy. Since y is going from -1, to 1, that integral is $\int_{-1}^1 (-y^4- 2y^2+ 1)dy$ .

Second problem:

Line integral of ydx + (x+y^2)dy where C is the ellipse 4x^2 + 9y^2 = 36, with counterclockwise rotation

This is a little harder because neither x nor y is a function of the other so we have to set up a "parameterization". Fortunately, there is a "standard" parameterization of an ellipse like this: x= 3cos(t), y= 2sin(t). Do you see why? 4x^2+ 9y^2= 4(9cos^2(t))+ 9(4sin^2(t))= 36(cos^2(t)+ sin^2(t))= 36.

Now, from x= 3cos(t), dx= -3 sin(t)dt and from y= 2sin(t), dy= 2cos(t)dt. Going completely around the ellpse, t goes from 0 to $2\pi$ .

**NickE** · July 28th 2012, 11:05 AM

Thanks that helps me understand it better. I have had instruction in this but I was a more general lecture. We only worked a few examples but it wasn't enough for me to fully grasp how to work any problem. The best way for me to learn anything is to do it repetitively and to do different kinds of examples.

**Prove It** · July 28th 2012, 08:36 PM

If you use the parameterisation given by HallsOfIvy, with $\displaystyle \begin{align*} x = 3\cos{t} \implies dx = -3\sin{t}\,dt \end{align*}$ and $\displaystyle \begin{align*} y = 2\sin{t} \implies dy = 2\cos{t}\,dt \end{align*}$ with $\displaystyle \begin{align*} 0 \leq t \leq 2\pi \end{align*}$ , then

$\displaystyle \begin{align*} \oint_C{\left[ y\,dx + \left(x + y^2\right)dy \right]} &= \int_0^{2\pi}{\left\{ 2\sin{t}\left(-3\sin{t}\,dt\right) + \left[ 3\cos{t} + \left(2\sin{t}\right)^2 \right]\left( 2\cos{t}\,dt \right) \right\} } \\ &= \int_0^{2\pi}{ -6\sin^2{t}\,dt + \left(6\cos^2{t} + 8\cos{t}\sin^2{t}\right) dt } \\ &= \int_0^{2\pi}{\left(6\cos{2t} + 8\cos{t}\sin^2{t}\right)dt } \\ &= \left[ 3\sin{2t} + \frac{8}{3}\sin^3{t} \right]_0^{2\pi} \\ &= 0 \end{align*}$

If you use Green's Theorem

$\displaystyle \begin{align*} L &= y \implies \frac{\partial L}{\partial y} = 1 \\ M &= x + y^2 \implies \frac{\partial M}{\partial x} = 1 \end{align*}$

So

$\displaystyle \begin{align*} \oint_C{ \left[ y\,dx + \left(x + y^2\right)dy \right] } &= \int{\int_D{\left(1 - 1 \right)dx}\,dy} \end{align*}$

We have the ellipse $\displaystyle \begin{align*} 4x^2 + 9y^2 &= 36 \end{align*}$ . If we use the Change of Variables $\displaystyle \begin{align*} x = \frac{u}{2} \end{align*}$ and $\displaystyle \begin{align*} y = \frac{v}{3} \end{align*}$ , the Jacobian is

$\displaystyle \begin{align*} \frac{\partial \left(x, y \right)}{\partial \left(u, v \right)} &= \left| \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x }{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{matrix} \right| \\ &= \left| \begin{matrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{3} \end{matrix} \right| \\ &= \frac{1}{6} \end{align*}$

and the ellipse transforms to the circle given by

$\displaystyle \begin{align*} 4x^2 + 9y^2 &= 36 \\ 4\left(\frac{u}{2}\right)^2 + 9\left(\frac{v}{3}\right)^2 &= 36 \\ u^2 + v^2 &= 36 \end{align*}$

and the integral gets transformed to

$\displaystyle \begin{align*} \int{\int_D{0\,dx}\,dy} &= \int{\int_S{0\cdot \frac{1}{6}\,du}\,dv} \\ &= \int{\int_S{0\,du}\,dv} \end{align*}$

And the region, now being a circle, should be transformed to polar coordinates. When this happens, we have the bounds $\displaystyle \begin{align*} 0 \leq r \leq 6 \end{align*}$ and $\displaystyle \begin{align*} 0 \leq \theta \leq 2\pi \end{align*}$ , which gives

$\displaystyle \begin{align*} \int_0^{2\pi}{\int_0^6{0\cdot r\,dr}\,d\theta} &= \int_0^{2\pi}{\int_0^6{0\,dr}\,d\theta} \\ &= \int_0^{2\pi}{\left[C_1\right]_0^6 d\theta} \\ &= \int_0^{2\pi}{0\,d\theta} \\ &= \left[C_2\right]_0^{2\pi} \\ &= 0 \end{align*}$

**HallsofIvy** · July 29th 2012, 12:18 PM

The original question was about doing the line integrals. Yes, the itegral around a closed curve can be done as a double integral but I don't think that was what was being asked.

Oh, and when you have $\int_A\int 0 dydx$ , no matter how complicated A is, I hardly think you need to convert to a simpler region! 0 has integral 0 over any region!

**Prove It** · July 29th 2012, 07:14 PM

Originally Posted by HallsofIvy

The original question was about doing the line integrals. Yes, the itegral around a closed curve can be done as a double integral but I don't think that was what was being asked.

Oh, and when you have $\int_A\int 0 dydx$ , no matter how complicated A is, I hardly think you need to convert to a simpler region! 0 has integral 0 over any region!

I was writing in response to replies by the poster, saying he wasn't sure about the bounds etc. And yes, I know that the integral over 0 is 0, I was showing what process needs to be followed in any case.

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