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I've been trying to work out how to change the order of integration (my tutor requested both ways - this is not a homework assignment) for the following problem when calculating using Green's Theorem:

$\oint\limits_C (6y - 3y^2 + x)~ \delta x + yx^3~ \delta y$ where C is shown in the image below.

I believe the correct solution is: $\oint\limits_{C} (6y - 3y^2 + x) \delta~ x + yx^3 \delta~ y = \int\int\limits_D \frac{\delta Q}{\delta x} - \frac{\delta P}{\delta y}~ \delta A = \int\limits_0^1\int\limits_{3x}^{5x} 3yx^2 - 6 + 6y~ \delta y~ \delta x = 14.8$

I am trying to calculate the same integral but with the order of integration the other way round (dx first then dy but can't seem to find the right limits that yield the same solution.

I believe the limits for the outer integral is 0 to 5 and that the inner integral involves using y/5 and y/3 but I'm not sure how to do this.

Hope that makes sense and thanks in advance for your help!

enter image description here

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If you are changing the order and integrating wrt $dx$ first, then you need to split your integral into two.

i) First between $0 \leq y \leq 3$, notice that $x$ is bound between lines $y = 5x$ and $y = 3x$.

So $\frac{y}{5} \leq x \leq \frac{y}{3}$

ii) Now between $3 \leq y \leq 5$, note that the lower bound of $x$ continues to be defined by the line $y = 5x$ but the upper bound is defined by line $x = 1$.

So the second integral will have bounds $3 \leq y \leq 5, \frac{y}{5} \leq x \leq 1$.

Can you take it from here? Let me know if any questions.

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    $\begingroup$ I see. I thought it would be able to be done in a single integral. I see the reasoning for splitting into two is that the upper bound changes for the two intervals for y. Thank you so much!! $\endgroup$ Mar 6, 2021 at 14:50
  • $\begingroup$ You are welcome! $\endgroup$
    – Math Lover
    Mar 6, 2021 at 14:50

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