# Application of integration

1. Oct 20, 2013

### Dumbledore211

1. The problem statement, all variables and given/known data

Show that the area of the plane bounded by the hyperbola (x)^1/2+(y)^1/2=(a)^1/2 and the axes of the coordinates is a^2/6

2. Relevant equations
∫yd x

3. The attempt at a solution
I just can't integrate the function (a)^1/2-(x)^1/2 considering the fact that there are x as well as a and should I take the limit from 0 to a for x??????

2. Oct 20, 2013

### haruspex

There's nothing stopping you integrating that, but it would not be ∫ydx. What is y equal to?
The limits should be at the axes, i.e. where x or y is zero. Does that correspond to x = 0 or a?

3. Oct 20, 2013

### Dumbledore211

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

Last edited: Oct 20, 2013
4. Oct 20, 2013

### haruspex

Start by expanding the 'squared'.

5. Oct 22, 2013

### Dumbledore211

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??

Last edited: Oct 22, 2013
6. Oct 22, 2013

### haruspex

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.

7. Oct 23, 2013

### Dumbledore211

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???

8. Oct 23, 2013

### haruspex

You missed a term when you integrated.

9. Oct 23, 2013

### Dumbledore211

How come I missed a term when I integrated??? Besides the integration of a becomes 0 with respect to x. Hence, the negative sign should also be taken into account because I expanded {(a)^1/2-(x)^1/2}^2

10. Oct 23, 2013

### haruspex

No, the derivative of a constant would be zero, but not the integral.

11. Oct 26, 2013

### Dumbledore211

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

12. Oct 26, 2013

### haruspex

I don't see a second part. It asks you to find an area, which you have now done.

13. Oct 26, 2013

### Dumbledore211

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...

Last edited: Oct 26, 2013
14. Oct 26, 2013

### haruspex

Let's not start a competition over who does the stupidest things. I might win.