yuki
  • yuki
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Sorry, I'm half here, half not...distractions.
anonymous
  • anonymous
So you don't know how these integrals are put together?
yuki
  • yuki
It's been 7years since I took my last calculus class ;p

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More answers

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

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