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, 11 months 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 \(\dfrac{30^{\circ}}{360^{\circ}} \cdot \pi \cdot \left((20\sqrt{6})^2 - (20\sqrt{3})^2 \right) = \boxed{100\pi}\)

Jimmy Kariznov - 4 years, 11 months ago

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Comment deleted Jul 28, 2013

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

Michael Tang - 4 years, 11 months ago

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Pi Han Goh - 4 years, 11 months ago

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

Michael Tong - 4 years, 11 months ago

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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.

Andrea Marino - 4 years, 11 months ago

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