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Find area of the region bounded by the circular arc in 1st Quadrant

  1. Apr 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the area of the region in the first quadrant, which is bounded by the x-axis, the line x = 2 and the circular arc x^2 + y^2 = 8


    2. Relevant equations



    3. The attempt at a solution

    I didn't use the hint given in the question but does my answer still makes sense. Did I set up the integral correctly? (My attempt is on the attachment)
     

    Attached Files:

    Last edited by a moderator: Apr 4, 2014
  2. jcsd
  3. Apr 3, 2014 #2

    phinds

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    I found it highly advisably in math to learn to do quick/simple calculations where possible to check the sanity of more complex calculations. It is trivially easy in this case. Takes about 2 seconds.

    What is the area of the entire circle? What is the portion of that area that is in one quadrant. How does that figure compare with your answer? How should it compare with your answer?

    This technique won't always be convincing that your answer is right or wrong but in this particular case I'd say it IS convincing that your answer is either (1) obviously very close to what has to be right or (2) obviously wrong. You do the math :smile:
     
  4. Apr 3, 2014 #3
    Ok. So the area of the circle is 8pi. The area of the square inside the circle would be 16. Subtracting that from each other would give 8pi - 16 to give the area inside the circle not containing the square. We want the 1st quadrant so (8pi - 16)/4 = 2pi - 4. The square in the first quadrant is 2 x 2 = 4. The area inside the 1st quadrant excluding the square is 2pi - 4. But the area asked would also include the area inside the circle and larger than y =2. This the area inside the 1st quadrant is 4 + pi - 2 = pi + 2

    Kinda took longer than 2 seconds. But I guess it works.
     
  5. Apr 3, 2014 #4

    phinds

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    Yeah, I don't know why you did what you did, but it definitely is the hard way. Just do exactly what I said. Take the area of the whole circle, which is, trivially, 8pi. Take 1/4 of that which is 2pi. Compare that (about 6.3) to your answer of pi + 2 (about 5.2) and you see that your answer is a bit less than the area of the circle portion in that quadrant, which is just what it should be. Took about 2 seconds. WAY longer to explain than to do.
     
  6. Apr 4, 2014 #5

    Curious3141

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    From your sketch, it looks like you're finding the wrong area.

    The area you're finding is bounded by the y-axis (x = 0), x = 2 and the arc.

    The area they want you to find is bounded by the x-axis (y = 0), x = 2 and the arc.

    EDIT: also, from the attachment, you're expected to use polar coordinates. You don't seem to be doing so.
     
    Last edited by a moderator: Apr 4, 2014
  7. Apr 4, 2014 #6
    It also says to find the area in the 1st quadrant. Which implies that the area is bounded by the x-axis, y-axis and the circular arc.

    The only way I can use polar coordinates is by setting theta with the range [pi/4 , pi/2] from x = 0 to x = 2. Then I add a second integral which can be solved by the line y = x from x = o to x = 2.

    However, I still don't know any way to use the integral with 1/cos^2(u).

    Do you(or anybody) know how?
     
  8. Apr 4, 2014 #7

    Curious3141

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    Then pray tell what the area in my attachment is supposed to represent?

    Let the circle's radius (##\sqrt{8}##) be represented by ##R##.

    There are three ways (at least) to find this area:

    1) Use Cartesian coordinates and evaluate ##\displaystyle \int_2^R\sqrt{R^2-x^2}dx ##

    2) Use Polar coordinates and evaluate ##\displaystyle \int_0^{\frac{\pi}{4}}\frac{1}{2}R^2 d\theta ## then subtract off the area of the right triangle with vertices at ##\displaystyle (0,0)## , ##\displaystyle (2,0)## and ##\displaystyle (2,2)##.

    3) Use Polar coordinates and directly evaluate the double integral ##\displaystyle \int_0^{\frac{\pi}{4}} \int_{\frac{2}{\cos\theta}}^R r dr d\theta##

    All methods give the same answer. The second way is the easiest by far, giving a very quick answer. The last way is the only one that uses that hint, and it's presumably the one they want you to use. Can you see why the integral is set up the way it is?
     

    Attached Files:

  9. Apr 4, 2014 #8
    Oh wow...I guess I really screwed up. Thanks for clarifying.
     
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