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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... :(

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  1. anonymous
    • 5 years ago
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    Sorry, I'm half here, half not...distractions.

  2. anonymous
    • 5 years ago
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    So you don't know how these integrals are put together?

  3. Yuki
    • 5 years ago
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    It's been 7years since I took my last calculus class ;p

  4. anonymous
    • 5 years ago
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    It's similar to finding the area under the curve, \[y=F(x)\]for \[a \le x \le b\]

  5. Yuki
    • 5 years ago
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    take your time, I'm the one who's getting help.

  6. anonymous
    • 5 years ago
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    We basically go about this through substitution.

  7. anonymous
    • 5 years ago
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    We know that x = f(t), and so there will be alpha, beta such that a=f(alpha), b=f(beta). Also, \[dx=f'(t)dt\]

  8. anonymous
    • 5 years ago
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    So\[A=\int\limits_{a}^{b}H(x)dx=\int\limits_{f(\alpha)}^{f(\beta)}H(f(t))f'(t)dt\]

  9. Yuki
    • 5 years ago
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    by H(x) do you mean F(x) ?

  10. anonymous
    • 5 years ago
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    Since y=H(x), y=H(f(t))=g(t) here, so\[A=\int\limits_{f(\alpha)}^{f(\beta)}g(t)f'(t)dt\]

  11. anonymous
    • 5 years ago
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    Yeah, sorry...H=F here...sloppy accounting.

  12. anonymous
    • 5 years ago
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    We're assuming the curve is traced out at most once as t roams from its initial value to its last.

  13. anonymous
    • 5 years ago
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    Fine so far?

  14. anonymous
    • 5 years ago
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    Okay...I'll assume yes...and continue...

  15. Yuki
    • 5 years ago
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    are you saying,\[\int\limits_{x(0)}^{x(2a)} y(t)*x'(t) dt\]

  16. anonymous
    • 5 years ago
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    \[g(t)=2a \cos ^2 t\]and\[f(t)=2a \tan t \rightarrow f'(t)=2a \sec^2 t\]

  17. anonymous
    • 5 years ago
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    yes

  18. anonymous
    • 5 years ago
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    I'm used to my notation... :[

  19. Yuki
    • 5 years ago
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    let me try using your notation,

  20. Yuki
    • 5 years ago
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    is it

  21. anonymous
    • 5 years ago
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    \[\therefore g(t)f'(t)=4a^2 \cos^2 t \sec^2 t=4a^2\]

  22. anonymous
    • 5 years ago
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    The integral is now easy. You have to find the limits.

  23. Yuki
    • 5 years ago
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    \[\int\limits_{0}^{\pi/4} 2acos^2(t)*2\sec^2(t) dt ?\]

  24. anonymous
    • 5 years ago
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    Yes, except you left out an 'a'.

  25. Yuki
    • 5 years ago
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    that was very clear explanation right there. bravo

  26. anonymous
    • 5 years ago
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    So it's\[4a^2\int\limits_{0}^{\pi/4}dt\]

  27. anonymous
    • 5 years ago
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    Easy!

  28. Yuki
    • 5 years ago
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    so the answer is

  29. anonymous
    • 5 years ago
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    pi a^2

  30. Yuki
    • 5 years ago
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    pia^2

  31. anonymous
    • 5 years ago
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    Concur...

  32. Yuki
    • 5 years ago
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    awesome. I will try to solve it on my own again.

  33. anonymous
    • 5 years ago
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    Great :)

  34. Yuki
    • 5 years ago
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    oh, one question.

  35. Yuki
    • 5 years ago
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    does volume work similarly ?

  36. anonymous
    • 5 years ago
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    You would have to make suitable substitutions again.

  37. Yuki
    • 5 years ago
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    gotcha

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