Double Integrals in Polar Coordinates - Example 2

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  • เผยแพร่เมื่อ 24 ม.ค. 2025

ความคิดเห็น • 32

  • @Yoaedn
    @Yoaedn 9 ปีที่แล้ว +7

    this guy is awesome. he goes at the perfect pace. some videos out there are way too slow.

  • @lfcforever
    @lfcforever 11 ปีที่แล้ว

    Many thanks mate... you are surely preparing me for my exam with your thorough steps :)

  • @djjcyxz
    @djjcyxz 8 ปีที่แล้ว

    Your videos are saving my Calc III class. I will definitely will recommend it to my friends!

    • @Mathispower4u
      @Mathispower4u  8 ปีที่แล้ว

      +Julio C. S Glad I could help. Thank you for the recommendations.

  • @fawzyhegab
    @fawzyhegab 11 ปีที่แล้ว

    great video , I wonder what is the name of the programe to get this calculator you have used through the video ?

  • @safaaaz5512
    @safaaaz5512 11 ปีที่แล้ว +1

    Thanks for your videos ,very helpful as usual and btw i like the quotes at the end, really inspiring. God bless you.

  • @kushagrasharma3900
    @kushagrasharma3900 8 ปีที่แล้ว +1

    U just saved my grades there mate....thnx😊

  • @Name-pn5rf
    @Name-pn5rf 8 ปีที่แล้ว +3

    Why does the limit of theta vary from 0 to ¥/2? Shouldn't it be from 0 to ¥? (¥=pi)

    • @michelec6468
      @michelec6468 7 ปีที่แล้ว +2

      My question exactly....

    • @derekpoon5308
      @derekpoon5308 6 ปีที่แล้ว +1

      Theta is measured from origin, not from center of circle.

  • @rmwhite13691
    @rmwhite13691 13 ปีที่แล้ว

    I keep getting burned by these types of problems where the lower bound of integration for the internal integral is the line y = x. How do I convert this equation and solve for r? The only answer I come up with is r = r*tan(theta), which just turns out to be garbage since you're now trying to evaluate the partial integral by replacing an r with another r.

  • @CocoGras
    @CocoGras 8 ปีที่แล้ว

    Maybe I'm missing something, but 1/24 seems kind of small for the area of half of a circle with radius of 1/2. If area of a circle is pi*radius

  • @terekingli
    @terekingli 11 ปีที่แล้ว

    why isn't the limit of r not 0 to 1/2 considering that 1/2 is the radius of the region?

  • @SN-th1bo
    @SN-th1bo 10 ปีที่แล้ว

    Mathispower4u Thank you for this video- just wondering why theta is equal to 90 or pi/2 and not pi? When I graph it in rectangular just like the previous example we were able to get theta from the graph. Thank you

    • @Mathispower4u
      @Mathispower4u  10 ปีที่แล้ว

      Because r = cos(theta) and when theta = 0, we are at the point (1,0) and when theta is pi/2, we are at the point (0,0). Therefore the semicircle is traced from r = 0 to r = pi/2.

    • @TheJq32
      @TheJq32 8 ปีที่แล้ว +1

      But how do we know that we are traveling from the point (1,0) to (0,0) if we are not allowed graphing calculators on the test?

  • @johnvalgon5660
    @johnvalgon5660 7 ปีที่แล้ว

    why we use r=0 to r=cos(theta) instead of r=one half ? it is just a semicircle

  • @znhait
    @znhait 11 ปีที่แล้ว

    Well, y = x passes through the angle pi/4. if it was a circle centered at the origin r would go from 0 to whatever the radius is.

  • @piemaster4
    @piemaster4 14 ปีที่แล้ว

    from 4:08-5:13 I just used the previous formula you had x^2+y^2=x to get r^2=rcostheta. then it works the same :)

  • @12Nowhere
    @12Nowhere 12 ปีที่แล้ว

    at 6:04, '' and when theta is 90 degrees or pie/2 we are back at the origin'' ???i
    when theta is 180 degrees or pie we are back at the origin.
    whe theta is 90 degrees we are at ( 1/2,1/2).

  • @DalerAsrorov
    @DalerAsrorov 11 ปีที่แล้ว +1

    Thanks a lot. Keep the videos going!

  • @eddywangchang
    @eddywangchang 8 ปีที่แล้ว +6

    this is better than patrickJMT and khan

  • @NeverLieToYa
    @NeverLieToYa 11 ปีที่แล้ว

    the point is the sketch is in the first quadrant which is 90 degrees..so it'll back at its origin :)

  • @1995sanchezbaybee
    @1995sanchezbaybee 9 ปีที่แล้ว

    Love your videos!!!

  • @khanqadeer8347
    @khanqadeer8347 2 ปีที่แล้ว

    Your are intelligence man.

  • @Conta_Google
    @Conta_Google 6 ปีที่แล้ว

    Thanks for your time sr!!!

  • @C00kieSandwiches
    @C00kieSandwiches 13 ปีที่แล้ว

    i wish you were my teacher.

  • @sabrinacamargo529
    @sabrinacamargo529 6 ปีที่แล้ว

    I found it easier to replace the function that's being integrated with the square root of r^2 rather than plug in all that.
    r^2 = x^2 + y^2

  • @jngibson24
    @jngibson24 13 ปีที่แล้ว

    You are amazing..

  • @darcash1738
    @darcash1738 หลายเดือนก่อน

    we see that x is from 0 to 1, placing it somewhere in the first or third quadrants. and that y has a sqrt at the top, placing it in the first or second. Combining these facts, it is solely in the first.
    thus we get theta bounds: [0, pi/2].
    Then we see that y has lower bound 0, upper bound sqrt(x-x^2).
    y = 0
    y = sqrt(x-x^2)
    --> sqrt(x-x^2) = 0
    x - x^2 = 0
    r(costheta - r) = 0
    --> r = 0, r = costheta.
    these are the bounds for our r: [0, costheta]

  • @Beaverphant
    @Beaverphant 13 ปีที่แล้ว

    @rmwhite13691 Same problem here

  • @samsonexe2088
    @samsonexe2088 13 ปีที่แล้ว

    Thank You Very Much!