x = 2a*tan(t) and y = 2a*cos^2(t) what is the area in the first quadrant bounded by the above parametric equations, x=0 and x=2a? I know how to find an area under some curves in rectangular and polar coordinates, but not for parametric curves... :(

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x = 2a*tan(t) and y = 2a*cos^2(t) what is the area in the first quadrant bounded by the above parametric equations, x=0 and x=2a? I know how to find an area under some curves in rectangular and polar coordinates, but not for parametric curves... :(

Mathematics
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Sorry, I'm half here, half not...distractions.
So you don't know how these integrals are put together?
It's been 7years since I took my last calculus class ;p

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It's similar to finding the area under the curve, \[y=F(x)\]for \[a \le x \le b\]
take your time, I'm the one who's getting help.
We basically go about this through substitution.
We know that x = f(t), and so there will be alpha, beta such that a=f(alpha), b=f(beta). Also, \[dx=f'(t)dt\]
So\[A=\int\limits_{a}^{b}H(x)dx=\int\limits_{f(\alpha)}^{f(\beta)}H(f(t))f'(t)dt\]
by H(x) do you mean F(x) ?
Since y=H(x), y=H(f(t))=g(t) here, so\[A=\int\limits_{f(\alpha)}^{f(\beta)}g(t)f'(t)dt\]
Yeah, sorry...H=F here...sloppy accounting.
We're assuming the curve is traced out at most once as t roams from its initial value to its last.
Fine so far?
Okay...I'll assume yes...and continue...
are you saying,\[\int\limits_{x(0)}^{x(2a)} y(t)*x'(t) dt\]
\[g(t)=2a \cos ^2 t\]and\[f(t)=2a \tan t \rightarrow f'(t)=2a \sec^2 t\]
yes
I'm used to my notation... :[
let me try using your notation,
is it
\[\therefore g(t)f'(t)=4a^2 \cos^2 t \sec^2 t=4a^2\]
The integral is now easy. You have to find the limits.
\[\int\limits_{0}^{\pi/4} 2acos^2(t)*2\sec^2(t) dt ?\]
Yes, except you left out an 'a'.
that was very clear explanation right there. bravo
So it's\[4a^2\int\limits_{0}^{\pi/4}dt\]
Easy!
so the answer is
pi a^2
pia^2
Concur...
awesome. I will try to solve it on my own again.
Great :)
oh, one question.
does volume work similarly ?
You would have to make suitable substitutions again.
gotcha

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