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My math book has a question of changing order of the following integral:

$$int_0^aint_{sqrt{ax-x^2}}^{sqrt{ax}} f(x,y),dx,dy.$$

So basically I have to change $dxdy$ into $dydx$ and their respective limits.

This is an example which is provided in my math textbook related to the problem. (For showing you diagram and make it easier for me to explain my confusion further.)

Image Link

So in my problem, while changing the order of integration, the resultant changed integral order is coming in 3 terms (say, IOnfoot1+IOnfoot2+IOnfoot3). I1 is for OLC, I2 is for AMC and I3 is for BLCM.(Please refer to the diagram, which I shared in the form of Image)

While changing the circle, i.e. $y=sqrt{ax-x^2}$ into form of $x$, we get $x=frac{1}{2}left(apmsqrt{a^2-4y^2}right)$.

I tried to solve it ahead but I stucked on a problem of deciding in ± which to choose to get higher (upper) limit and to get lower limit. But in the answersheet of my Textbook, the following is given:-

In OLC the limit of y changes into [ from y=0 to y=a/2 ] and limit of x changes into [ x= y^2/a to x= {a-sqrt(a^2-4y^2) }/2]
Which means {a-sqrt(a^2-4y^2)}/2 is greater than {a+sqrt(a^2-4y^2}/2

In AMC, again we see that {a+sqrt(a^2-4y^2}/2 is taken as lower limit of x.

So, I could not understand how to decide or check which number will be greater and which number will be smaller between {a-sqrt(a^2-4y^2)}/2 is and {a+sqrt(a^2-4y^2}/2

My questions are:-

(i) How can we testify that which number can be taken as upper limit and which number can be taken as lower limit of x in the different terms while changing the order of integration, in accordance with the given question.



(ii) Is there any method to test whether one algebraic expression (or number) is greater than the other in Mathematics? If yes, then please tell me.

P.S.:- Thank You So Much in advance if you willing to amswer my question. Thanks again if you edited my question to make math expressions easily understable, as I don't know how to do it here.

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The question in the textbook is stated incorrectly. The original order of integration is $dydx$ NOT $dxdy$. This is because we are integrating $y$ first on the inner integral and then $x$. It cannot be $dxdy$ because we cannot integrate with respect to $x$ and have the limits of the integral be written in terms of $x$.

i) Typically we integrate on the $x$ axis from left to right. By doing so we can see that the graph $y =sqrt{2ax} $ is the leftmost curve. Therefore the lower limit will be $x=frac{y^2} {2a} $. Similarly, moving rightwards along the $x$ axis we 'hit' the left section of the circular region next. For the curve $(x-a)^2 + y^2 = a^2$ we can rearrange to get $x = a{+/-} {sqrt{a^2-y^2}} $. This curve takes the positive sqrt for the right half of the circle and the negative sqrt for the left half of the circle. As we need the equation of the left half of the circle, we need to take $x = a-sqrt{a^2-y^2} $ for the upper limit. This does not imply that this value is greater than the other.

ii) In general taking the positive sqrt will always result in a greater number than the negative sqrt, but when integrating, we do not always want the greater value, we want the equation of the curve we are using to integrate.