Solving Line Integrals

[burt]

This is the problem I was given and my work so far. I feel like I set up my integral right, but did I convet the equation to a parametric one in an appropiate way?
All I did was convert the equation for $c$ and then plug the parametric equations in for $x$ and $y$ in the integrand - multiplying by the arc length which is $\sqrt{x^2+y^2}$. Is this right?

 

[burt]

After looking over my work, I found one error - I replaced $t^2$ with $t$ which changed my problem. But, my question about the set-up still remains.
This is my work should really look like: $$\text{Define }x^2+y^2=16\text{ parametrically: } x=t\text{ and }y=\sqrt{16-t^2}\\\int_c(x^2+y^2)ds=\int^4_{-4}(t^2+16-t^2)(\sqrt{1+\frac{1}{4(16-t^2)}})dt\\=\int^4_{-4}(16)(\sqrt{1+\frac{1}{4(16-t^2)}}dt\\=131.96\text{ using CAS}$$
I also was thinking that maybe I should double my answer to make up for the fact that I really have $y=\pm\sqrt{16-t^2}$. This makes my solution: $263.921$.
The question is the same as I had in the previous post - this is just correcting what I saw as an error.

 

[tkhunny]

1) Why would you multiply by 2 when you are given y >= 0?

2) I don't see how you are following the ENTIRE contour?

3) Are you SURE you want to use the algebraic parameterization? For the circular portion, I would think trigonometric parameterization might simplify your life.

 

[burt]

Why would you multiply by 2 when you are given y >= 0

I see - I should just leave it once. So it's not really a half circle then...

 

[burt]

Are you SURE you want to use the algebraic parameterization? For the circular portion, I would think trigonometric parameterization might simplify your life.

For my parameters I should use $x=4\sin(t)$ and $y=4\cos(t)$ then? I see how that now traces out the entire circle.
What was wrong with my other one? Making it a square root which got rid of the negative?

 

[burt]

Using my new parameters, I got $$\int_{c}\left(x^2+y^2\right)ds=\int_{-4}^{4}\left(16\left(\sin^2{\left(t\right)}+\cos^2{\left(t\right)}\right)\right)\left(\sqrt{16\sin^2{\left(t\right)}+16\cos^2{\left(t\right)}}\right)dt$$
This simplifies to: $$=\int_{-4}^{4}\left(16\right)\left(4\right)dt$$
Which equals $512$.
This makes a lot more sense, but I'm still trying to understand where I went wrong with the other answer. It seems that something went wrong with my arc length.

 

[tkhunny]

I see - I should just leave it once. So it's not really a half circle then...

I'll give you my #2. I was thinking you hadn't traced the diameter on the x-axis, making it a closed figure.

 

[burt]

I'll give you my #2. I was thinking you hadn't traced the diameter on the x-axis, making it a closed figure.

Okay
Did I correct my answer? Or am I still making a mistake?

 

[burt]

@tkhunny
I don't mean to bother you - I just appreciate your advice, and the way you made me work through the problem. Did I follow through all the way?
I've been realizing that there is another problem - I have to change my limits of integration, don't I?
Changing my limits of integration to $0$ and $\pi$ would result in a final answer of $64\pi$

 

[topsquark]

Note that $$x^2 + y^2 = r^2$$ in polar coordinates. So your original integral becomes
$$\int_C (x^2 + y^2) ~ ds = \int_0^{ \pi } ~ r^2 ~ (r ~ d \theta ) = r^3 \int_0^{ \pi } d \theta$$ (since r = 4 is a constant.)

-Dan

 

[burt]

Note that $$x^2 + y^2 = r^2$$ in polar coordinates. So your original integral becomes
$$\int_C (x^2 + y^2) ~ ds = \int_0^{ \pi } ~ r^2 ~ (r ~ d \theta ) = r^3 \int_0^{ \pi } d \theta$$ (since r = 4 is a constant.)

-Dan

I see. That results in the same answer $64\pi$, right?

 

[topsquark]

I see. That results in the same answer $64\pi$, right?

Yes.

-Dan