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yuki
 5 years ago
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... :(
yuki
 5 years ago
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, I'm half here, half not...distractions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you don't know how these integrals are put together?

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0It's been 7years since I took my last calculus class ;p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's similar to finding the area under the curve, \[y=F(x)\]for \[a \le x \le b\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0take your time, I'm the one who's getting help.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We basically go about this through substitution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We 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
 5 years ago
Best ResponseYou've already chosen the best response.0So\[A=\int\limits_{a}^{b}H(x)dx=\int\limits_{f(\alpha)}^{f(\beta)}H(f(t))f'(t)dt\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0by H(x) do you mean F(x) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since y=H(x), y=H(f(t))=g(t) here, so\[A=\int\limits_{f(\alpha)}^{f(\beta)}g(t)f'(t)dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, sorry...H=F here...sloppy accounting.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We're assuming the curve is traced out at most once as t roams from its initial value to its last.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...I'll assume yes...and continue...

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0are you saying,\[\int\limits_{x(0)}^{x(2a)} y(t)*x'(t) dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[g(t)=2a \cos ^2 t\]and\[f(t)=2a \tan t \rightarrow f'(t)=2a \sec^2 t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm used to my notation... :[

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0let me try using your notation,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\therefore g(t)f'(t)=4a^2 \cos^2 t \sec^2 t=4a^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The integral is now easy. You have to find the limits.

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\pi/4} 2acos^2(t)*2\sec^2(t) dt ?\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, except you left out an 'a'.

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0that was very clear explanation right there. bravo

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So it's\[4a^2\int\limits_{0}^{\pi/4}dt\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0awesome. I will try to solve it on my own again.

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0does volume work similarly ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You would have to make suitable substitutions again.
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