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- Homework Statement
- Show that ##\pmb{\int_C x^2y\; ds = \frac{1}{3}}## where ##C## is the circular arc in the first quadrant of the unit circle.

- Relevant Equations
- (1) The line element ##ds^2=dx^2+dy^2\Rightarrow ds = \sqrt{1+\left( \frac{dy}{dx} \right)^2} dx## in cartesian coordinates.

(2) The line element ##ds = a d\phi##, where ##a## is the radius of the circle, in polar coordinates. (Here ##a=1\Rightarrow ds = d\phi##).

(3) The equation of the quadrant is : ##x^2 + y^2 = 1## with ##0\le (x,y) \le 1##

**Problem :**We are required to show that ##I = \int_C x^2y\;ds = \frac{1}{3}##.

**Attempt :**Before I begin, let me post an image of the problem situation, on the right. I would like to do this problem in

__three ways__, starting with the simplest way - using (plane) polar coordinates.

**(1) In (plane) polar coordinates**: Remembering that the radius of the (circular) quadrant is 1 (##a=1##), we have the replacements : ##x = \cos \phi\,\, , y = \sin \phi\;\; , ds = d\phi##. The limits of the integral vary from ##\phi = 0\rightarrow \pi/2##, where (I assume) that the quadrant is navigated in an anti- clockwise manner.

Hence the line integral ##I = \int_{0}^{\pi/2} \cos^2 \phi \sin\phi \; d\phi##. Choosing ##z=\cos\phi## makes ##dz = -\sin \phi d\phi## and the limits going from 1 to 0. Our integral becomes ##I = - \int_1^0 z^2 dz = \int_0^1 z^2 dz = \frac{z^3}{3}\Biggr|_0^1 = \boxed{\frac{1}{3}}##, which yields the answer easily.

##\boxed{\text{This is where I run into a small problem}}##

**(2) Using cartesian coordinates with**The line integral ##I = \int_C x^2 y \; ds##. My aim is to convert everything inside the integral sign into the variable ##x##. Clearly, ##y = \sqrt{1-x^2}##, from the equation of the (circular) quadrant. As for ##ds^2=dx^2+dy^2\Rightarrow ds = \sqrt{1+\left( \frac{dy}{dx} \right)^2} dx##. Using ##y=\sqrt{1-x^2}\Rightarrow \frac{dy}{dx} = \frac{-x}{\sqrt{1-x^2}}##. This makes sense because the slope of the ##ds## is negative in the first quadrant (see small black line in picture representing the direction of ##ds##).

*x*as the integral variable :Thus ##1+\left (\frac{dy}{dx} \right)^2 = 1+\frac{x^2}{1-x^2} = \frac{1}{1-x^2}##. Thus, we have ##ds = \frac{1}{\sqrt{1-x^2}}dx\Rightarrow yds = dx##.

Putting the information from above, the integral ##I = \int_{x=1}^{x=0} x^2 dx = \boxed{-\frac{1}{3}}## (which is wrong by the minus ##-## sign}.

**A hint or help would be welcome.**

##\boxed{\text{Curiously, things work out fine if I use cartesian coordinates treating y as the integral variable}}##

**(3) Using cartesian coordinates with**The line integral ##I = \int_C x^2 y \; ds##. My aim is to convert everything inside the integral sign into the variable ##y##. Clearly, ##x = \sqrt{1-y^2}\Rightarrow x^2 = 1-y^2##, from the equation of the (circular) quadrant. As for ##ds^2=dx^2+dy^2\Rightarrow ds = \sqrt{1+\left( \frac{dx}{dy} \right)^2} dy##. Using ##x=\sqrt{1-y^2}\Rightarrow \frac{dx}{dy} = \frac{-y}{\sqrt{1-y^2}}##. Thus ##\left( 1+\frac{dx}{dy} \right)^2 = 1+\frac{y^2}{1-y^2} = \frac{1}{1-y^2}##. Thus, we have ##ds = \frac{dy}{\sqrt{1-y^2}}##.

*y*as the integral variable :The integral ##I = \int_0^1 \sqrt{1-y^2}ydy##. Putting ##1-y^2 = z^2\Rightarrow -ydy = zdz##, which yields ##I = -\int_1^0 z^2 dz = \boxed{\frac{1}{3}}##, again yielding the required answer.

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