Results 1 to 7 of 7

Thread: Find b s.t. y=b divides area between y=25-x^2, y=0 into 2 equal halves

  1. #1

    Find b s.t. y=b divides area between y=25-x^2, y=0 into 2 equal halves

    1. LarCalcET6 7.1.069.

    Find b such that the line y = b divides the region bounded by the graphs of y = 25 - x2 and y = 0 into two regions of equal area.




    I got:

    . . . . .[tex]\displaystyle \int_0^5\, (25\, -\, x^2)\, dx\, =\, \dfrac{250}{3}[/tex]

    Then tried to find:

    . . . . .[tex]\displaystyle \int_0^b\, (25\, -\, x^2)\, dx\, =\, \dfrac{125}{3}[/tex]

    But I get stuck when I try to evaluate it:

    . . . . .[tex]\displaystyle 3\int_0^b\, (25\, -\, x^2)\, =\, 125[/tex]

    . . . . .[tex]\displaystyle 3\left[\, 25x\, -\, \frac{1}{3}x^3\right]_0^b\, =\, 125[/tex]

    . . . . .[tex]\displaystyle 75b\, -\, b^3\, =\, 125[/tex]

    Sent from my LGLS755 using Tapatalk
    Last edited by stapel; 02-02-2018 at 03:08 PM. Reason: Typing out the text in the graphic; creating useful subject line.

  2. #2
    Senior Member
    Join Date
    Nov 2017
    Location
    Rochester, NY
    Posts
    1,586
    Quote Originally Posted by Seed5813 View Post
    1. LarCalcET6 7.1.069.

    Find b such that the line y = b divides the region bounded by the graphs of y = 25 - x2 and y = 0 into two regions of equal area.




    I got:

    . . . . .[tex]\displaystyle \int_0^5\, (25\, -\, x^2)\, dx\, =\, \dfrac{250}{3}[/tex]

    Then tried to find:

    . . . . .[tex]\displaystyle \int_0^b\, (25\, -\, x^2)\, dx\, =\, \dfrac{125}{3}[/tex]

    But I get stuck when I try to evaluate it:

    . . . . .[tex]\displaystyle 3\int_0^b\, (25\, -\, x^2)\, =\, 125[/tex]

    . . . . .[tex]\displaystyle 3\left[\, 25x\, -\, \frac{1}{3}x^3\right]_0^b\, =\, 125[/tex]

    . . . . .[tex]\displaystyle 75b\, -\, b^3\, =\, 125[/tex]
    You're using a line x=b, not y=b, to divide the region!

    Try again. There are a couple ways you might do this.
    Last edited by stapel; 02-02-2018 at 03:08 PM.

  3. #3
    Quote Originally Posted by Dr.Peterson View Post
    You're using a line x=b, not y=b, to divide the region!

    Try again. There are a couple ways you might do this.
    Y= 25-x^2
    y-25=x^2
    sqrt(25-y)=x

    Int_0^25 (25-y)^(1/2)dy
    U=25-y
    du=-dy
    -int_0^25(u^(1/2))du
    -[(2/3)u^(3/2)]_0^25
    -[(2/3)(25-y)^(3/2)]_0^25
    -[-(2/3)(25)^(3/2)]
    [(2/3)(125)]
    (250/3)

    Also how do I know when to use x and when to use y?

    Sent from my LGLS755 using Tapatalk

  4. #4
    Quote Originally Posted by Seed5813 View Post
    Y= 25-x^2
    y-25=x^2
    sqrt(25-y)=x

    Int_0^25 (25-y)^(1/2)dy
    U=25-y
    du=-dy
    -int_0^25(u^(1/2))du
    -[(2/3)u^(3/2)]_0^25
    -[(2/3)(25-y)^(3/2)]_0^25
    -[-(2/3)(25)^(3/2)]
    [(2/3)(125)]
    (250/3)

    Also how do I know when to use x and when to use y?

    Sent from my LGLS755 using Tapatalk
    ^wrong btw

    Sent from my LGLS755 using Tapatalk

  5. #5
    Sorry here's the whole thing:

    Y= 25-x^2
    y-25=x^2
    sqrt(25-y)=x

    Int_0^25 (25-y)^(1/2)dy
    U=25-y
    du=-dy
    -int_0^25(u^(1/2))du
    -[(2/3)u^(3/2)]_0^25
    -[(2/3)(25-y)^(3/2)]_0^25
    -[-(2/3)(25)^(3/2)]
    [(2/3)(125)]
    (250/3)

    Int_0^b ((25-y)^(1/2)dy=(250/3)
    U=25-y
    Du=-dy
    -int_0^b(u^(1/2))du=(250/3)
    -[(2/3)(25-y)^(3/2)]_0^b=(250/3)
    -[(2/3)(25-b)^(3/2) - (250/3)]=(250/3)
    -(2/3)(25-b)^(3/2) + (250/3) = (250/3)
    -(2/3)(25-b)^(3/2) = 0
    (25-b)^(3/2)=0
    25-b = 0
    B=-25

    Sent from my LGLS755 using Tapatalk

  6. #6
    Quote Originally Posted by Dr.Peterson View Post
    You're using a line x=b, not y=b, to divide the region!

    Try again. There are a couple ways you might do this.
    Found the problem

    Sent from my LGLS755 using Tapatalk

  7. #7
    Senior Member
    Join Date
    Nov 2017
    Location
    Rochester, NY
    Posts
    1,586
    Quote Originally Posted by Seed5813 View Post
    Sorry here's the whole thing:

    Y= 25-x^2
    y-25=x^2
    sqrt(25-y)=x

    Int_0^25 (25-y)^(1/2)dy
    U=25-y
    du=-dy
    -int_0^25(u^(1/2))du
    -[(2/3)u^(3/2)]_0^25
    -[(2/3)(25-y)^(3/2)]_0^25
    -[-(2/3)(25)^(3/2)]
    [(2/3)(125)]
    (250/3)

    Int_0^b ((25-y)^(1/2)dy=(250/3)
    U=25-y
    Du=-dy
    -int_0^b(u^(1/2))du=(250/3)
    -[(2/3)(25-y)^(3/2)]_0^b=(250/3)
    -[(2/3)(25-b)^(3/2) - (250/3)]=(250/3)
    -(2/3)(25-b)^(3/2) + (250/3) = (250/3)
    -(2/3)(25-b)^(3/2) = 0
    (25-b)^(3/2)=0
    25-b = 0
    B=-25
    You're working only with the right half of the region, but that doesn't affect the answer.

    The trouble here is that you didn't take half of the area (125/3), so your answer turned out to be actually b=+25, which is the whole region.

    I presume you realized that and got the correct answer.

    There's a formula for the area of a segment of a parabola: 2bh/3. But you were, of course, expected to do what you did.

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •