You should be able to solve this

Here you go

You failed calc iii didnt you?

I actually got an A in every class in the calc series, but I go to an engineering school and the math classes are easy. Literally never seen a cycloid before and I will be a certified engineer in a year.

What brainlet school do you go to?

You found the fucking integral over a rectangle, not over the cycloid.

could someone solve a related problem or something because i'm actually blanking on how to do this

my first instinct is to do dx=a(1-cos(t)dt and dy=asin(t)dt but then the integral wouldn't make sense

You'll be an engineering graduate but not a certified engineer. Big difference.

did you get asked this before or after recess?

Thats the thing. The parameteric equation forms the upper bound only. You need x to go from 0 to 2api, and y from 0 to the upper bound. But it's hard to make y a function of x - it's possible but you will get a massive function full of arc cosines and stuff. Not even wolfram can find the integral.

Oh yeah don't certified engineers have to take that really hard exam for certification?

there's two exams, one you can take before you graduate and the other you need to work as a junior engineer for some time before you can take it. the last one takes something like 12 hours. if you pass both you become a professional engineer.

Veeky Forums Should be able to solve this using Green's Theorem too.

something like this?

The idea is good, but you fucked up along the way. What does "integral from 0 to y of y^2 dy" mean? You need to be more pedantic.

The solution should be (35*pi*a^4)/12

I was supposing that the upper bound for the area on the y axis is the value of y at any point, and x is bounded between 0 and 2*pi*a which depends on the variable t so then i just make a variable shift for the limits of the integral, where was i wrong in doing that? Are you op? Are you sure the answer is that? How would you solve it?

wow you guys are retarded
[math]a = r , t = \theta[/math]
then integrate[math]dxdy = rdrd\theta[/math]

so rewrite x,y using a and t to find the boundaries. Then integrate that shit.

i meant rewrite a, t with x, y.

[math]t - 2tsin(t) - 2cos(t)=
\frac{1}{a^2} -2[/math] and [math]t - sin(t) = 1 - cos(t)[/math]

>where was i wrong in doing that? Are you op? Are you sure the answer is that? How would you solve it?

You cant get y^3 as a result of definite integral over y. Also you cant get cos(t) as a result of definite integral over t.

I'm OP, I checked the result in the textbook. I'd do it similarly to you, except I would take 0 to phi(x) as a boundry for y, where phi(x) is a (probably not elementary) function of x such that for every x where x=a(t-sint), phi(x)=a(1-cost), for t in [0, 2pi].

Then substitute x=a(t-sint), you get phi(t)=a(1-cost) and dx=a(1-cost)dt

Then it is just definite integration

Then do it, brainlet
Show your work

Ok, i got it, the problem was exactly what you said, i took the wrong limits for t when i did the inverse of x(t), i cannot do that, i shoud have done the following:
For x=0=a(t-sint) and x=2*pi*a=a(t-sint) and solve for t getting the boundries you said, t=0 and 2*pi, and as for the integral of y^2 i should have got (1/3)*y^3, thanks man

a is likely a constant, judging on the graph

either way that's plain out wrong, you'd need to do a jacobian transformation matrix

If "a" were a variable, you'd have
dx/da * dy/dt - dx/dt * dy/da
= a * a*sin(t) - a(1-cos(t)) * a
= a^2(sin(t) + cos(t)-1)
=> dxdy = (a^2(sin(t) + cos(t)-1))dadt

Which school do you go to in which doing CalcIII in the first semester is a regular thing? At my uni, there's only ~5-8 freshman in CalcIII.

are you drunk or something?

There are way more than 5-8 freshmen in Vector Calculus at a college like UT Austin or UCLA.

>12 hours
holy shit am I glad I am comp sci.

This is very simple, holy shit. Post in the next time.

Then solve it brainlet
Show you're work

> you're work
> you are work
brainlet

You can do this using Green's theorem.
Integrate y^3/3 around the contour.
The bottom portion is just 0,

dx = a(1 - cos t) dt

[math] \int_0^{2*\pi} (1/3)*a*(1 - \cos[t])^3*(a*(1 - \cos[t])) dt [/math]

Result: [math]35 \pi a^2 / 12 [/math]

Your a brainlet

Finally someone not a brainlet

>Result: 35πa^2/12
Oh wait youre still a brainlet.
Learn to exponentiate, fag.

...

Shit, got the first line wrong. Brb, killing myself.

Yeah, so it becomes
[eqn]\frac{a^3}{3} \int_0^{2\pi} \left( 1- cos(t) \right)^4 \space dt = \frac{35 \pi}{4} \cdot \frac{a^3}{3} = \frac{35\pi a^3}{12}[/eqn]

Nice work, senpai

OwO