I've been trying to teach myself a few modules of my university course in preparation before I start and I'm currently attempting some Vector Calculus, but I'm finding it much more difficult to grasp than previous modules. I was beginning to think I was getting the hang of it until I came across the question below.

The closed curve $C$ in the $z=0$ plane consists of the arc of the parabola $y^2=4ax$ $(a>0)$ between the points $(a,±2a)$ and the straight line joining $(a, ∓2a)$. The area enclosed by $C$ is $A$. Show, by calculating the integrals explicitly, that $$\int_C(x^2y\,\mathrm d\mkern1mu x + xy^2\,\mathrm d\mkern1mu y)=\int_A(y^2−x^2)\,\mathrm d\mkern1mu A=\frac{104}{105}a^4$$ where $C$ is traversed anticlockwise.

Apologies if my attempts here are laughable, at the moment I'm just trying to familiarise myself with a few of the concepts so that I'll have already met it when it comes to studying it in my course. As a result of this I'm rushing quite quickly through the topic and not worrying so much about my understanding of 'why' but just trying to compute a few answers. It's likely that I'll throw around some incorrect terms here and there, this is just so that people can see my thought process.

What I'm having issues with:

For $\int_C(x^2y\,\mathrm d\mkern1mu x + xy^2\,\mathrm d\mkern1muy)$ I thought what we could do is use the line integral formula $\int_{C} \mathbf f\cdot d\mathbf r = \int_{t_0}^{t_1} \mathbf f(\mathbf r)\cdot \mathbf r'(t)\ dt$, with the parametrisation ${\bf r}(t)=(\frac{t^2}{4a},t)$ we have $\int^{2a}_{-2a}(x^2y,xy^2)\cdot (\frac t{2a},1) dt$ which when we substitute in $t$ for $x$ and $y$ and evaluate the integral I get something like $\frac{152}{35}a^4$ which is clearly wrong. I doubt highly that I've made an arithmetic error, it's far more likely that I've fundamentally misunderstood and got mixed up, so apologies for that.

I get $\int_A(y^2−x^2)\,\mathrm d\mkern1mu A=\int^{2a}_{-2a}\int^{\frac{y^2}{4a}}_0(y^2-x^2) \mathrm d\mkern1mux \mathrm d\mkern1muy$ and I won't even bother going further becuase I know this is already wrong.

My question (finally): Answer the original question, but also if possible give criticism of my attempts, why am I wrong, what have I mixed up, where would my answer be right? I'm also unclear as to what the last line of the question is saying "where $C$ is traversed anticlockwise" (I have an idea of what this could mean, but not why it would ever affect an answer).

Apologies again for my embarrassing attempts (thankfully this is anonymous), I know it's a big ask, (essentially reteach me an entire topic) so any input is appreciated.

Thank you


There is nothing to be ashamed about here. You are trying to learn something on your own and got pretty close. The only mistake you made was in the interpretation of what you were already given. If you take a function $f = (f^1,f^2)$ and $c(t) = (x = x(t), y = y(t))$ and let $C$ be some curve over which the line integral makes sense and parametrized over $[a,b]$ then;

$$\int_C f \cdot ds = \int_{t=a}^b \langle f^1(c(t)),f^2(c(t))\rangle \cdot \langle x'(t),y'(t) \rangle = \int_{t=a}^b \left(f^1(c(t)) x'(t) + f^2(c(t)) y'(t)\right) \ dt = \int_C f^1 dx + f^2 dy$$

So the form of the line integral you have is the same as the standard one. What you have to plug in for $x,y$ are the parametrized coordinate functions in $r(t)$ and $dx,dy$ just means take the derivatives of these coordinate functions. To not deprive you of the problem, I just did a few computations to show you what I mean. I hope this helps.


$(1)$ You have $r(t) = \left(x = \dfrac{t^2}{4a},y=t\right)$ where $t \in [-2a,2a]$

$(2)$ Then $dx = \dfrac{t}{2a}$, $dy = 1$ and $x^2y = \dfrac{t^6}{16a^2}$, $xy^2 = \dfrac{t^4}{4a} $

  • $\begingroup$ Thanks for the response. Sorry I'm going to continue to be an idiot here, but I can't see the difference between our approaches. I had $\int^{2a}_{-2a}(x^2y,y^2x)\cdot (\frac t{2a},1)dt$ which when we substitute in $t$ we get $\int^{2a}_{-2a}(\frac{t^6}{16a^2},\frac{t^4}{4a})\cdot (\frac t{2a},1)dt$ Surely this is also what you're suggesting would yield? Any thoughts on the other integral? $\endgroup$ – Aka_aka_aka_ak Sep 9 '16 at 17:06

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