I get two different answers, ##a^2## and ##0.5a^2##, by using two different methods. Which is the correct answer?(adsbygoogle = window.adsbygoogle || []).push({});

The family of curve for ##y^2=4u(u-x)## is given by the blue curves, and that for ##y^2=4v(v+x)## is given by the red curves.

Method 1:

Evaluate the integral ##I## directly in the xy-plane.

##\int_0^adx\int_0^{\sqrt{4a(a-x)}}dy\,y(x^2+y^2)^{-1/2}##

##=\int_0^adx[(x^2+y^2)^{1/2}]_0^{\sqrt{4a(a-x)}}##

##=\int_0^adx\big((x^2+4a^2-4ax)^{1/2}-|x|\big)##

##=\int_0^adx\big(|x-2a|-|x|\big)##

##=\int_0^adx\big((2a-x)-x\big)##

##=[2ax-x^2]_0^a##

##=a^2##

Method 2:

Evaluate the integral ##I## in the uv-plane. (The blue curves are ##u=1##, ##u=2##, etc. The red curves are ##v=1##, ##v=2##, etc.)

First, we express ##x## and ##y## in terms of ##u## and ##v##:

##x=u-v##

##y=2\sqrt{uv}##

The above can be easily verified by substituting into ##y^2=4u(u-x)## and ##y^2=4v(v+x)##.

##y=-2\sqrt{uv}## is rejected as the integral is to be done over the region ##y>0##.

##I=\frac{1}{4}\int_0^adv\int_0^aduJ\frac{2\sqrt{uv}}{u+v}##, where ##J## is the Jacobian.

The factor ##\frac{1}{4}## is present because the integral is to be done only over the top left quadrant, but ##\int_0^adv\int_0^aduJ## covers the area of the pointed ellipse that extends over all quadrants.

Pointed ellipse (Its left side is the red curve ##v=a## and its right side is the blue curve ##u=a##):

The pointed ellipse is symmetrical about the x and y axes. Thus the area of the pointed ellipse in each quadrant is the same.

##J=\frac{u+v}{\sqrt{uv}}##

##I=\frac{1}{4}\int_0^adv\int_0^adu2=0.5a^2##

Which is the correct answer?

EDIT: In order for ##I=a^2##, ##\int_0^adv\int_0^aduJ## must be covering only ##\frac{1}{2}## the area of the pointed ellipse. But why? Has it got to do with the rejection of ##y=-2\sqrt{uv}##?

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# I Change of variables in double integral

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