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# Square root transformation

My friend shared this problem with me and I thought it was very interesting.

A transformation of a first quadrant maps the points $$(x, y)$$ to $$( \sqrt{x}, \sqrt{y})$$. The vertices of a quadrilateral ABCD are $$A(900, 300), B(1800, 600), C(600, 1800), D(300, 900).$$ Find the area of its image under the said transformation.

Note by Michael Tong
4 years ago

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Line $$AB$$ is $$y = \dfrac{1}{3}x$$, which maps to $$y^2 = \dfrac{1}{3}x^2$$, i.e. $$y = \dfrac{1}{\sqrt{3}}x$$.

Line $$BC$$ is $$x+y = 2400$$, which maps to $$x^2+y^2 = 2400$$.

Line $$CD$$ is $$y = 3x$$, which maps to $$y^2 = 3x^2$$, i.e. $$y =\sqrt{3}x$$.

Line $$DA$$ is $$x+y = 1200$$, which maps to $$x^2+y^2 = 1200$$.

Convert all of these to polar to get: $$\theta = 30^{\circ}$$, $$r = 20\sqrt{3}$$, $$\theta = 60^{\circ}$$, $$r = 20\sqrt{6}$$.

Hence, the new region is exactly a $$30^{\circ}$$ sector of a doughnut with inner radius $$20\sqrt{3}$$ and outer radius $$20\sqrt{6}$$.

Therefore, the area is $$\dfrac{30^{\circ}}{360^{\circ}} \cdot \pi \cdot \left((20\sqrt{6})^2 - (20\sqrt{3})^2 \right) = \boxed{100\pi}$$ · 4 years ago

Comment deleted Jul 28, 2013

All points are shifted with the transformation - not just the vertices. So, the image is NOT a quadrilateral. · 4 years ago

ARH! MY BAD! · 4 years ago

As the other Michael T said, every point of the quadrilateral is shifted. That's the tricky part of the question. · 4 years ago

Unluckily I could not write the complete solution for time reasons, but i post the four main observations which bring to the solution:

1. If $$S$$ is a closed region of the plane and $$(x,y) \in S$$, it holds $$(x',y') \in S'$$; so the area of $$A'B'C'D'$$ is the area of the region delimited by $$A'B'$$, $$B'C'$$, $$C'D'$$, $$D'A'$$.

2. A line is transformed in an elliptical or hyperbolical arc; infact if $$(x,y)$$ satisfies $$ax+by+c=0$$, then $$(x',y')$$ satisfies $$ax'^2+by'^2+c=0$$. With this process we can clearly identify the equation of the elliptical or hyperboloic arc in which a line is trasformed.

3. An ellipse can be obtained by stretching a circle. If $$C$$ is a circle of ray $$R$$, and $$E$$ is an ellipse of semi-axis $$R, \lambda R$$ (respectively vertical and horizontal), every region $$S \in C$$ of area $$A$$ is stretched in a region $$S' \in E$$ of area $$\lambda A$$. Furthermore, an horizontal stretch clearly does not modifiy vertical coordinates.

4. Unluckily, I don't know any trick to compute areas bounded by hyperbolas; I would use integral calculus.

Using this facts we can (with a little bit of dirty work xD) evaluate the area of A'B'C'D' by summing and subtracting certain areas. I'm pretty sure there is a more elegant method which provide a relation between areas for the square root trasformation. I will search it. · 4 years ago