How Do I Evaluate the Integral \(\int(z^2+1)^2dz\) Over a Cycloid Path?

[Stumped1]

$$\int(z^2+1)^2dz$$
Evaluate this over the cycloid$$x=a(\theta-sin\theta)$$ and $$y=a(1-cos\theta)$$ for $$\theta =0$$ to $$\theta = 2\pi$$

Am I on the right track, or do I need to approach this a different way?

for $$z^2$$ we have $$(x+iy)^2$$, so $$x^2-y^2 + i2xy$$ for the real part.

$$a^2(\theta-sin\theta)-a^2(1-cos\theta)^2$$
$$a^2[\theta^2-1 -2sin\theta + sin^2\theta -1 + 2cos\theta - \cos^2\theta ]$$

Aside from taking $$-cos^2\theta = sin^2\theta -1$$ and consolidating the $$sin^2\theta$$

Not sure what else I can do to simplify this.

Thanks for any help!

 

[Prove It]

$$\int(z^2+1)^2dz$$
Evaluate this over the cycloid$$x=a(\theta-sin\theta)$$ and $$y=a(1-cos\theta)$$ for $$\theta =0$$ to $$\theta = 2\pi$$

Am I on the right track, or do I need to approach this a different way?

for $$z^2$$ we have $$(x+iy)^2$$, so $$x^2-y^2 + i2xy$$ for the real part.

$$a^2(\theta-sin\theta)-a^2(1-cos\theta)^2$$
$$a^2[\theta^2-1 -2sin\theta + sin^2\theta -1 + 2cos\theta - \cos^2\theta ]$$

Aside from taking $$-cos^2\theta = sin^2\theta -1$$ and consolidating the $$sin^2\theta$$

Not sure what else I can do to simplify this.

Thanks for any help!

Is your contour even closed?

 

[Stumped1]

Is your contour even closed?

Seems that it is since it will be one rotation of the cycloid, since it is parameterized by $$0 \leq\theta \leq 2\pi$$

 

[Prove It]

If the contour is closed and your function doesn't have any singular points (which this doesn't) then the integral is equal to 0 by Cauchy's Theorem.

However I am not so sure that this is closed... See here...

 

[Stumped1]

If this is not closed, continue w/ my approach?

 

[Prove It]

If this is not closed, continue w/ my approach?

My understanding is that all complex integration requires having a closed contour, so if this is what you have been given I would assume that it is closed. I'd advise you to check with your lecturer to be sure.

 

[chisigma]

$$\int(z^2+1)^2dz$$
Evaluate this over the cycloid$$x=a(\theta-sin\theta)$$ and $$y=a(1-cos\theta)$$ for $$\theta =0$$ to $$\theta = 2\pi$$

Am I on the right track, or do I need to approach this a different way?

for $$z^2$$ we have $$(x+iy)^2$$, so $$x^2-y^2 + i2xy$$ for the real part.

$$a^2(\theta-sin\theta)-a^2(1-cos\theta)^2$$
$$a^2[\theta^2-1 -2sin\theta + sin^2\theta -1 + 2cos\theta - \cos^2\theta ]$$

Aside from taking $$-cos^2\theta = sin^2\theta -1$$ and consolidating the $$sin^2\theta$$

Not sure what else I can do to simplify this.

Thanks for any help!

The fact that f(z) is analytic means that its integral along a path depends only from the starting and ending points. Here f(z) is anayltic, so that the integral is...

$\displaystyle \int_{C} f(z)\ d z = \int_{0}^{2\ \pi\ a} (1+x^{2})^{2}\ dx\ (1)$

The details are left to You...

Kind regards

$\chi$ $\sigma$

 

[alyafey22]

$$\int(z^2+1)^2dz$$
Evaluate this over the cycloid$$x=a(\theta-sin\theta)$$ and $$y=a(1-cos\theta)$$ for $$\theta =0$$ to $$\theta = 2\pi$$

Am I on the right track, or do I need to approach this a different way?

for $$z^2$$ we have $$(x+iy)^2$$, so $$x^2-y^2 + i2xy$$ for the real part.

$$a^2(\theta-sin\theta)-a^2(1-cos\theta)^2$$
$$a^2[\theta^2-1 -2sin\theta + sin^2\theta -1 + 2cos\theta - \cos^2\theta ]$$

Aside from taking $$-cos^2\theta = sin^2\theta -1$$ and consolidating the $$sin^2\theta$$

Not sure what else I can do to simplify this.

Thanks for any help!

Hint : the function you are integrating has an anti-derivative in $\mathbb{C}$.