How do I integrate for the Area below a one-to-two curve?

A

Al-Layth

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what I mean by one-to-two curve is that the curve has 2 y value outputs for a single x value input.

As an example of this problem I have attached an image to illustrate what I mean. The equation of that circle is
[math]9 = (x-5)^2 +(y-5)^2[/math]and I am trying to formulate an integral to compute the greyed area. (no synthetic geometric approach allowed)

so naturally I figured I should make y the subject and do the definite integral and got:
[math]y=\sqrt{9-(x-5)^2}+5[/math]
My point of confusion however is: which y value will this give ?
the higher one or the lower one? if it is the higher one, then in my integral the area computed will include both the entire circle and the grey area.
if its the lower one well thast great but how do I make sure I have found the correct y(x) relation that will only give me the lowest of the y values in a one-to-multiple cartesian curve? thanks
(note: the image is not a problem I am just using it to illustrate my question)




areabelow1-2relation.jpg
 
The lower half of the circle is [math]y=-\sqrt{9-(x-5)^2}+5[/math]The area bounded between the curve and the x-axis is
[math]A = \int_{2}^8 -\sqrt{9-(x-5)^2}+5 ~ dx[/math]
 
If y^2 = anything (no y's on the right hand side), then y = (+/-)sqrt(anything).
You need to decide whether to use the +, - or maybe both
 
Last edited:
Al-Layth said:
I figured I should make y the subject and do the definite integral and got:
[math]y=\sqrt{9-(x-5)^2}+5[/math]
My point of confusion however is: which y value will this give ?
the higher one or the lower one? if it is the higher one, then in my integral the area computed will include both the entire circle and the grey area.
if its the lower one well thast great but how do I make sure I have found the correct y(x) relation that will only give me the lowest of the y values in a one-to-multiple cartesian curve? thanks
Click to expand...
When you solved for y, you forgot to use both square roots; those are the reason for the one-to-two nature of the function. Your solution should have been [math]y=\pm\sqrt{9-(x-5)^2}+5[/math]
It should be clear that the positive sign gives the higher point, and the negative sign gives the lower point.
 
Dr.Peterson said:
When you solved for y, you forgot to use both square roots; those are the reason for the one-to-two nature of the function. Your solution should have been [math]y=\pm\sqrt{9-(x-5)^2}+5[/math]
It should be clear that the positive sign gives the higher point, and the negative sign gives the lower point.
Click to expand...
Oh, that’s it?

Thank you so much i am the village idiot
 
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