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Application of integration

  • #1

Homework Statement



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

Homework Equations


∫yd x


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??????
 

Answers and Replies

  • #2
haruspex
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I just can't integrate the function (a)^1/2-(x)^1/2 considering the fact that there are x as well as a
There's nothing stopping you integrating that, but it would not be ∫ydx. What is y equal to?
should I take the limit from 0 to a for x??????
The limits should be at the axes, i.e. where x or y is zero. Does that correspond to x = 0 or a?
 
  • #3
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
 
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  • #4
haruspex
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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
Start by expanding the 'squared'.
 
  • #5
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??
 
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  • #6
haruspex
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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??
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
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
haruspex
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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???
You missed a term when you integrated.
 
  • #9
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
haruspex
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the integration of a becomes 0 with respect to x.
No, the derivative of a constant would be zero, but not the integral.
 
  • #11
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
haruspex
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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
I don't see a second part. It asks you to find an area, which you have now done.
 
  • #13
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...
 
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  • #14
haruspex
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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...
Let's not start a competition over who does the stupidest things. I might win.
 

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