I was working on a question that asked me to find the mass of a given curve using a specified density function. My initial approach was to use the line integral of the first kind, which intuitively seemed suitable for this question. However, upon completing the line integral, I decided to try an alternative approach using a normal double integral in polar coordinates. To my surprise, the two approaches yielded different answers.
This is the question: Find the mass of a wire in the shape of the semi-circle $$y = \sqrt{4 - x^2}$$ from $(2, 0)$ to $(−2, 0)$, and the density $\rho$ at a point $(x,y)$ on the wire is $\rho(x,y)=y^2$.
Approaches:
1st approach: (Line Integral of the First Kind) $$\int_{0}^{\pi} 4 \sin^2 \theta \sqrt{(-2 \sin \theta)^2 + (2 \cos \theta)^2} \, d\theta$$
$$Result = 4\pi$$
2nd Approach: (Double Integral with Polar Coordinates) $$\int\limits_{0}^{2} \int\limits_{0}^{\pi} r^3 \sin^2(\theta) \, d\theta \, dr$$ $$Result = 2\pi$$
Upon rechecking my calculations, I am unable to identify the source of the discrepancy between the two approaches. Aside from that, why would the double integral approach result in a lower mass compared to the line integral approach?