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The actual question states the following;
"Find the mass of a Quater-Disc (in terms of R), in the first quadrant, of radius 'R' if density varies as D = xy"

My first thought was somehow turning this problem into another one in hyperbolic coordinates and integrating for the transformation of a circle there with density varying radially, but then I realised I can't do that because I don't know how to.

I also thought about taking hyperbolic elements and the density function as xy=k where k goes from 0 to R/2 but I just couldn't figure out how to take elements that don't have an anchor for me to keep constant.

I can visually grasp the problem statement and its intention but I fail to materialize it into equations I can evaluate..

Any help is appreciated.

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Given the density function $D = xy$ and the quarter-disc in the first quadrant with radius $R$, we'll set up the integral to find the mass.$$\\$$Okay so mainly you can do that in 4 steps that you're gonna read right now, try to do that on your own for your own good

The first step: Parameterization:

  • Let $x$ vary from $0$ to $R$ and $y$ vary from $0$ to $\sqrt{R^2 - x^2}$ to describe the quarter-disc in the first quadrant.

The second step: Density Function:

  • $D = xy$.

The third step: Setting Up the Integral:

  • The mass ($M$) is given by the double integral over the quarter-disc: $$ M = \int_{0}^{R} \int_{0}^{\sqrt{R^2 - x^2}} xy \, dy \, dx $$

The fourth step: Evaluating the Integral:

  • First, integrate with respect to $y$ from $0$ to $\sqrt{R^2 - x^2}$, then integrate the result with respect to $x$ from $0$ to $R$.

Let's start by integrating $xy$ with respect to $y$ from $0$ to $\sqrt{R^2 - x^2}$: $$ \int_{0}^{\sqrt{R^2 - x^2}} xy \, dy = x \cdot \left[ \frac{1}{2}y^2 \right]_{0}^{\sqrt{R^2 - x^2}} = x \cdot \frac{1}{2} (R^2 - x^2) $$

Now, integrate this expression with respect to $x$ from $0$ to $R$: $$ M = \int_{0}^{R} x \cdot \frac{1}{2} (R^2 - x^2) \, dx $$

$$ M = \frac{1}{2} \int_{0}^{R} (R^2x - x^3) \, dx $$

$$ M = \frac{1}{2} \left[ \frac{1}{2} R^2x^2 - \frac{1}{4} x^4 \right]_{0}^{R} $$

$$ M = \frac{1}{2} \left( \frac{1}{2} R^2 \cdot R^2 - \frac{1}{4} R^4 \right) $$

$$ M = \frac{1}{2} \left( \frac{1}{2} R^4 - \frac{1}{4} R^4 \right) $$

$$ M = \frac{1}{2} \cdot \frac{1}{4} R^4 $$

$$ M = \frac{1}{8} R^4 $$

So, the mass of the quarter-disc in terms of $R$ is $ \frac{1}{8} R^4 $.

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  • $\begingroup$ You could also set it up in polar coordinates $$M = \iint_Q D\, rd\theta\,dr =\int_0^Rr^3\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta\,dr=\frac 12\int_0^Rr^3\,dr\int_0^{\pi/2}\sin 2\theta\,d\theta$$ $\endgroup$ Apr 22 at 20:19

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