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- Homework Statement
- Show that ##\boxed{\oint_C x^2 y \mathrm{d} s = -\frac{\sqrt{2}}{12}}## where ##C## is the closed triangle shown in the figure below.

- Relevant Equations
- This is an example of a "line integral" of the form ##\int_C \psi(x,y,z) ds##. (I am not sure what relevant equations to add here. I'd prefer to get right into my attempt of this question).

**As a part of the problem, the diagram shows the contour ##C##above on the left. The contour ##C## is divided into three parts, ##C_1, C_2, C_3## which make up the sides of the right triangle.
**

Problem statement :

Problem statement :

**Required to prove**: ##\boxed{\oint_C x^2 y \mathrm{d} s = -\frac{\sqrt{2}}{12}}##.

**Attempt :**We note straightaway that the line integral must be zero on lines ##C_1, C_3## as the values of ##x,y## are zero on ##C_1## and ##C_3## respectively. I require to find the line integral on the line ##C_2##.

The integral $$I = \int_{(1,0)}^{(0,1)} x^2y\, \mathrm{d}s$$ The line itself has the equation : ##x+y=1\Rightarrow y=1-x##. From the diagram above, ##s## and ##x## are oppositely directed along the line ##C_2##. Hence we have ##ds=-\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx##. The derivative ##dy/dx = -1\Rightarrow ds = -\sqrt{2} dx##.

Writing the integral I entirely in terms of ##x##, we get : $$\small{I = \int_{(1,0)}^{(0,1)} x^2y\, \mathrm{d}s = \int_1^0 -x^2(1-x)\sqrt{2} dx=\sqrt{2}\int_0^1 (x^2-x^3) dx= \sqrt{2} \left| \frac{x^3}{3} - \frac{x^4}{4}\right|_0^1= \sqrt{2}\left(\frac{1}{3}-\frac{1}{4}\right)=\boxed{\frac{\sqrt{2}}{12}}}$$

My answer is incorrect by the opposite sign. I must confess that I fail to see where I went wrong. Both variables ##x,y## are positive in the first quadrant where the integral is carried out. The contour ##C## is traversed in a counter-clockwise manner, which is positive by the right-hand screw rule. Still, any hint or help would be welcome.

I have a second matter regarding this problem. How can it be done using (plane) polar coordinates? I can write both ##x=\rho \cos\phi## and ##y=\rho\sin\phi## in polar coordinates, but I am confused as to how to write the line element ##\mathrm{d}s## in polar form. Still, any hint or help would be welcome.