# Area of a Quater-Circle with hyperbolic elements

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.

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 $$\frac{1}{8} R^4$$.

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