The problem involves finding the area bounded by a hyperbola defined by the equation (x)^(1/2) + (y)^(1/2) = (a)^(1/2) and the coordinate axes. The original poster attempts to show that this area equals a^2/6.
The discussion includes various attempts to integrate the function and clarify limits. Some participants provide hints and corrections regarding the integration process, while others express confusion about the negative results obtained during integration.
There is a mention of potential mistakes in integration and the need to consider the correct limits based on the axes. The original poster expresses uncertainty about the interpretation of the problem and the integration process.
There's nothing stopping you integrating that, but it would not be ∫ydx. What is y equal to?Dumbledore211 said:I just can't integrate the function (a)^1/2-(x)^1/2 considering the fact that there are x as well as a
The limits should be at the axes, i.e. where x or y is zero. Does that correspond to x = 0 or a?should I take the limit from 0 to a for x?
Start by expanding the 'squared'.Dumbledore211 said:y={ (a)^1/2- (x)^1/2}^2. The integration should be done with respect to x where y is just a function of x. Can you drop a hint as to how I should integrate it and is 0≤ x≤ a or 0≤ x≤ (a)^1/2 true
I already answered that. The other bounds are the axes, so the limits correspond to x = 0 for one and y = 0 for the other.Dumbledore211 said:Okay, then it becomes y= a-2(a)^1/2(x)^1/2+x. Now, integrating both sides we get y= - 4/3x^3/2(a)^1/2+x^2/2. But, what limits should I take? Is 0≤x≤a valid in this case??
You missed a term when you integrated.Dumbledore211 said:Taking the limit from 0 to a the integration with respect to x becomes y= -4a^2/3+a^2/2= -8a^2+3a^2/6= -5a^2/6 How come I have a negative sign and the area is not negative. Besides the answer is a^2/6. What mistakes am i making?
No, the derivative of a constant would be zero, but not the integral.Dumbledore211 said:the integration of a becomes 0 with respect to x.
I don't see a second part. It asks you to find an area, which you have now done.Dumbledore211 said:Sorry, I made a ridiculously silly mistake. Don't know what happened as I solved harder integration problems like this. How do I solve the second part of the question?? I have no clue as to how I can find out the axes of the coordinates to be a^2/6
Let's not start a competition over who does the stupidest things. I might win.Dumbledore211 said:Sorry for my foolishness. I completely misread the question at first I thought that I had to show that the area and the axes of the coordinates are a^2/6. Anyway, Thank you haruspex for bearing my ridiculousness and showing me the extent of silliness that I exhibit in solving the simplest problems. My brain malfunctions at times especially when I am doing a lot of math per day...