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Homework Help: Volume Around x=-r

  1. Jan 23, 2010 #1
    1. The problem statement, all variables and given/known data
    The region R enclosed by the curves x-2y=-2 and y=sqrt(x-2)+2 is rotated about the line x=-2. Find the volume of the resulting solid.

    2. Relevant equations
    [tex]$\displaystyle \Large V=\int _b^d A(y) dy$[/tex]
    [tex]$\displaystyle \Large V=\int _b^d\pi ((2 +x_R(y))^2-(2 +x_L(y))^2)dy.$[/tex]

    3. The attempt at a solution
    after making y1 = y2
    intersection points, (2, 2) and (6, 4).
    i need xR and xL, so
    xR = 2y-2
    xL = y2-4y+6

    [tex]$\displaystyle \Large V=\int _2^4\pi [(2 +(2y-2))^2-(2+(y^2-4y+6))^2]dy$[/tex]


    [tex]$\displaystyle \Large V=\int _2^4\pi [4y^2-(y^4 - 8y^3 +32y^2 +64y +64)]dy$[/tex]

    [tex]$\displaystyle \Large V=\pi [(-y^5)/5 + 2y^4 -(28y^3)/3 +32y -64y)]$[/tex]42

    V = 261.3805088

    what's wrong with my calculation?
  2. jcsd
  3. Jan 24, 2010 #2


    Staff: Mentor

    So adding in the extra 2 units, and renaming the right and left x values,
    xR = 2y
    xL = (y - 2)2 + 4

    It makes the work easier to leave in factored form -- fewer things to multiply and hence go wrong.
    I didn't check, but I think you might have gone astray in the line above.
    Try this integral:
    [tex]V=\pi \int_2^4 [(2y)^2 - ((y - 2)^2 + 4)^2][/tex]
    [tex]\Rightarrow V = \pi \int_2^4 [4y^2 - ((y - 2)^4 +8(y - 2)^2 + 16)][/tex]

    Don't expand the (y - 2) factors. They're easy enough to integrate without expanding them.

    Carrying out the integration, I get 224pi/15.
  4. Jan 24, 2010 #3
    oh okay.. i guess i messed up when i opened the brackets lol

    what about:
    The region R enclosed by the curves x-4 y=-31 and y=sqrt(x-1)+8 is rotated about the line y=7. Use cylindrical shells to find the volume of the resulting solid.

    there is only one intersection point.. or do i include the y=7 intersection too?
  5. Jan 24, 2010 #4


    Staff: Mentor

    No, there are two intersection points - (1, 8) and (17, 12).
  6. Jan 24, 2010 #5
    but its weird becuase when i graph it with a graphing calc i cant see the graphs intersecting at (1, 8) but anyways...

    i used this formula:
    [tex]$\displaystyle \Large V=\int _b^d S(y) dy=\int _b^d 2\pi(y-7)(x_R(y)-x_L(y))dy.$[/tex]

    and got
    [tex]$\displaystyle \Large V=\int _8^{12} 2\pi(y-7)(-31+4y-(y-8)^2 -1)dy.$[/tex]
    ==> [tex]$\displaystyle \Large V=\int _8^{12} 2\pi(y-7)(-y^2 +20y-96)dy.$[/tex]
    ==> [tex]$\displaystyle \Large V=\int _8^{12} 2\pi(-y^3 +13y^2 -236y+672)dy.$[/tex]
    ==>[tex]$\displaystyle \Large V= 2\pi[(-y^4)/4 +(13y^3)/3 -(236y^2)/2+672y]$[/tex]128

    and my final answer is V=35453.92029
    but its wrong.. is there something wrong with my calculation again?
  7. Jan 24, 2010 #6


    Staff: Mentor

    In the next line I get 27y^2, not 13y^2. Everything else looks good.
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