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ranyai12

  • 2 years ago

Evaluate the line integral 5ydx+2xdy where C is the straight line path from (4,4) to (7,8)

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  1. ranyai12
    • 2 years ago
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    I got 276 but was wrong

  2. ranyai12
    • 2 years ago
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    @Hero please help!

  3. SithsAndGiggles
    • 2 years ago
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    \[\int_C (5y~dx+2x~dy)\] The line connecting (4,4) and (7,8) is \(y=\dfrac{4}{3}x-\dfrac{4}{3}\) Parameterize \(C\) by \[\begin{cases}x=t\\ \\ y=\dfrac{4}{3}t-\dfrac{4}{3}\\ \\ 4<t<7 \end{cases}\] So you have \(dx=dt\) and \(dy=\dfrac{4}{3}dt\): \[\int_4^7 \bigg[5\left(\dfrac{4}{3}t-\dfrac{4}{3}\right)~dt+2t~\left(\dfrac{4}{3}dt\right)\bigg]\\ \int_4^7\bigg[\frac{28}{3}t-\frac{20}{3}\bigg]~dt\\ ~~~~~~~~~~~~~~~~\vdots\]

  4. ranyai12
    • 2 years ago
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    did you get 134?

  5. SithsAndGiggles
    • 2 years ago
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    Yep: http://www.wolframalpha.com/input/?i=Integrate%5B%2828%2F3%29*t-%2820%2F3%29%2C%7Bt%2C4%2C7%7D%5D Can't guarantee that's the right answer, but that's what my method is getting, so I'm sticking with it.

  6. ranyai12
    • 2 years ago
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    when i did it by hand, i ended up getting 288

  7. SithsAndGiggles
    • 2 years ago
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    Could you show your work? It might be a mistake in the integration, or the parameterization.

  8. ranyai12
    • 2 years ago
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    288 is wrong :(

  9. SithsAndGiggles
    • 2 years ago
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    Well what did you do when you tried it? How did you parameterize \(C\)?

  10. ranyai12
    • 2 years ago
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    no i integrated wrong! the answer is indeed 134

  11. SithsAndGiggles
    • 2 years ago
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    Oh ok then.

  12. ranyai12
    • 2 years ago
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    Thanks so much

  13. SithsAndGiggles
    • 2 years ago
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    You're welcome!

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