Area Under a Curve

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sandi2001

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So this question has been bugging me for several hours now. The problem is really simple until my prof wants me to find the area of the attached curve using HORIZONTAL STRIPS. Is it even possible to use horizontal strips in this type of curve? Can you guys help me out on how to approach this? Thank you very much!
 

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sandi2001 said:
So this question has been bugging me for several hours now. The problem is really simple until my prof wants me to find the area of the attached curve using HORIZONTAL STRIPS. Is it even possible to use horizontal strips in this type of curve? Can you guys help me out on how to approach this? Thank you very much!
Click to expand...
Sure, it's possible (though obviously not one's first choice). Just split the region into two parts (which look like they'll be equal), above and below the x-axis. Then you'll be using inverse trig functions.

Show us your work, so we can see if you're getting off to a bad start.
 
For horizontal strips, you will need to do it as two separate regions.

For y> 0 a horizontal strip will go from x= pi/2 to x= arcsin(y).
For y< 0 a horizontal strip will go from x= arccos(x) to pi.
 
Dr.Peterson said:
Sure, it's possible (though obviously not one's first choice). Just split the region into two parts (which look like they'll be equal), above and below the x-axis. Then you'll be using inverse trig functions.

Show us your work, so we can see if you're getting off to a bad start.
Click to expand...

Please let me know where I came off wrong. I'm really confused.
 

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HallsofIvy said:
For horizontal strips, you will need to do it as two separate regions.

For y> 0 a horizontal strip will go from x= pi/2 to x= arcsin(y).
For y< 0 a horizontal strip will go from x= arccos(x) to pi.
Click to expand...

Hello, HallsofIvy. Is this what you meant by the parameters? I'm really confused on how to use them. Should I use the parameters on the upper and lower bounds of the integral?
 

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There's one little problem: the range of arcsin is -pi/2 to pi/2, and your x has to be between pi/2 and pi. How can you adjust?

(Halls seems to have missed this; I probably would have, too, in giving a quick suggestion and expecting you to fill in any gaps.)
 
For A1 y goes from 0 to 1. That part is clear to you based on the your limits in your integral.

Now as pointed out by Dr Peterson, arcsin is defined between -pi/2 and pi/2. Since our y value is between 0 and 1 then the x is between 0 and pi/2. Is there another region of the graph that has x between 0 and pi/2 AND has the same area that you want?? The answer is yes. Just look at the graph and you'll see it.

Same thing for A2
 
Thank you very much everyone. With your thoughts, I realized on what part should I be focusing on the graph. Keep safe!
 
I've decided to use the curve under the parameters: 0 - pi/2 for arcsin, and just use the given arccos curve under x since it falls on the arccos range (0 to pi).
 

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sandi2001 said:
Can I turn A1 into pi/2 - arcsiny?
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I would define the inverse function in that region as pi - arcsin(y), which is how we find the angle in the second quadrant with sine y. That leads to exactly what you came out with.
 
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