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Vectors and calculus, lots of help needed (very )

  1. Nov 1, 2009 #1

    rock.freak667

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    Vectors and calculus, lots of help needed (very urgent)

    1. The problem statement, all variables and given/known data
    Let S be part of the cylinder x2+z2=1 that lies above the rectangle in the plane z=0 that has vertices (1/2,1/2,0),(1/2,-1/2,0),(-1/2,-1/2,0) and (-1/2,1/2,0). By evaluating ∫∫S 1 ds, find the surface area of S.


    3. The attempt at a solution
    For the cylinder x=cosθ ,z=sinθ

    r(θ,y)=cosθi+yj+sinθk

    0≤θ≤2π and -1/2≤y≤1/2


    rθ = -sinθi+cosθk

    ry=j

    rθxry=-sinθk-cosθj

    |rθxry|=1

    ∫∫S 1 ds = ∫∫|rθxry| dA


    [tex]\int_0 ^{2\pi} \int_{\frac{-1}{2}} ^{\frac{1}{2}} y dy d\theta[/tex]

    is this correct? (I am not getting the correct answer)





    1. The problem statement, all variables and given/known data
    Evaluate the surface integral ∫∫S (y2i+x2j+z4k).n ds

    where S is the part of the cone defined by [itex]z=4 \sqrt{x^2+y^2}, 0 \leq z \leq8 \and \ y \geq 0[/itex]. Assume S has an upward orientation.



    3. The attempt at a solution

    I know I need to find out what is n but I don't know what to parameterize to find n. :confused:



    1. The problem statement, all variables and given/known data

    Let E be the solid region enclosed by the paraboloid z=1-x2-y2 and the plane z=0. Evaluate the surface integral

    ∫∫(yi+xj+zk).n ds

    Where S is the boundary surface of E that has outward orientation.

    3. The attempt at a solution
    Evaluate 1. The problem statement, all variables and given/known data
    Let S be part of the cylinder x2+z2=1 that lies above the rectangle in the plane z=0 that has vertices (1/2,1/2,0),(1/2,-1/2,0),(-1/2,-1/2,0) and (-1/2,1/2,0). By evaluating ∫∫S 1 ds, find the surface area of S.


    3. The attempt at a solution
    For the cylinder x=cosθ ,z=sinθ

    r(θ,y)=cosθi+yj+sinθk

    0≤θ≤2π and -1/2≤y≤1/2


    rθ = -sinθi+cosθk

    ry=j

    rθxry=-sinθk-cosθj

    |rθxry|=1

    ∫∫S 1 ds = ∫∫|rθxry| dA


    [tex]\int_0 ^{2\pi} \int_{\frac{-1}{2}} ^{\frac{1}{2}} y dy d\theta[/tex]

    is this correct? (I am not getting the correct answer)





    1. The problem statement, all variables and given/known data
    Evaluate the surface integral ∫∫S (y2i+x2j+z4k).n ds

    where S is the part of the cone defined by [itex]z=4 \sqrt{x^2+y^2}, 0 \leq z \leq8 \and \ y \geq 0[/itex]. Assume S has an upward orientation.



    3. The attempt at a solution

    I know I need to find out what is n but I don't know what to parameterize to find n. :confused:



    1. The problem statement, all variables and given/known data

    Let E be the solid region enclosed by the paraboloid z=1-x2-y2 and the plane z=0. Evaluate the surface integral

    ∫∫(yi+xj+zk).n ds

    Where S is the boundary surface of E that has outward orientation.

    3. The attempt at a solution



    1. The problem statement, all variables and given/known data
    Evaluate ∫∫∫E 1/2 (x2+y2)2 dV

    Where E= {(x,y,z)|x2+y2≤4,-2≤z≤2}


    2. Relevant equations



    3. The attempt at a solution

    I tried to convert to cylindrical coordinates and got this

    0≤r≤2 0≤θ≤2π and -2≤z≤2


    [tex]\int \int \int \frac{1}{2}(x^2+y^2)^2 dV = \int_0 ^{2\pi} \int_0 ^2 \int_2 ^2 \frac{1}{2} \times 16 rdzdrd\theta[/tex]


    Once again, I get the wrong answer. Please help me.
     
  2. jcsd
  3. Nov 1, 2009 #2

    LCKurtz

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    Re: Vectors and calculus, lots of help needed (very urgent)

    For your first problem, the cylinder above the z = 0 plane, note that the portion of the cylinder above the given rectangle is not described by letting [itex]\theta[/itex] go from 0 to [itex]2\pi[/itex]. The correct values of [itex]\theta[/itex] should fix your problem.
     
  4. Nov 1, 2009 #3

    rock.freak667

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    Re: Vectors and calculus, lots of help needed (very urgent)

    I am measuring [itex]\theta[/itex] in the wrong way, so it should be from pi to 0 then?
     
  5. Nov 1, 2009 #4

    LCKurtz

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    Re: Vectors and calculus, lots of help needed (very urgent)

    For your cone consider parameterizing with cylindrical coordinates

    [tex]\vec{R}(r,\theta)[/tex]
    and proceeding like your first problem.
     
  6. Nov 1, 2009 #5

    LCKurtz

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    Re: Vectors and calculus, lots of help needed (very urgent)

    No. Draw a cross section, a circle of radius 1 above a line segment (-1/2,1/2) and figure out the values of [itex]\theta[/itex].
     
  7. Nov 1, 2009 #6

    rock.freak667

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    Re: Vectors and calculus, lots of help needed (very urgent)

    Which plane am I looking at the circle in, the xy or xz plane?

    If in the xz plane, how would I see the line segment?
     
  8. Nov 1, 2009 #7

    LCKurtz

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    Re: Vectors and calculus, lots of help needed (very urgent)

    If you looked at the end of your cylinder down the y-axis your cylinder would look like a circle and you would be looking at the edge of your rectangle [-1/2,1/2] on the x axis. Draw that picuture showing your corresponding values of [itex]\theta[/itex].
     
  9. Nov 1, 2009 #8

    LCKurtz

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    Re: Vectors and calculus, lots of help needed (very urgent)

    For your solid bounded by the paraboloid you could do it directly parameterizing the paraboloid with cylindrical coordinates and the xy plane in polar coordinates. Or you could apply the divergence theorem and do a volume integral, again, probably in cylindrical coordinates.
     
  10. Nov 1, 2009 #9

    LCKurtz

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    Re: Vectors and calculus, lots of help needed (very urgent)

    In your last problem, the volume integral, you have to substitute the appropriate variables for the expression:

    [tex](x^2 + y^2)^2[/tex]

    in terms of r. You can't just plug in 2 for r because on the inside of the volume r isn't equal 2 and your triple integral goes over the whole volume. And your inner integral goes from -2 to 2.
     
  11. Nov 1, 2009 #10

    LCKurtz

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    Re: Vectors and calculus, lots of help needed (very urgent)

    And I also just notice in your first problem, what you have written as

    ∫∫S 1 dS = ∫∫|rθxry| dA

    should be written as

    ∫∫S 1 dS = ∫∫|rθxry| dθdy

    When you use that formula you shouldn't put in an extra y.

    Good luck. I have to hit the sack here.
     
  12. Nov 2, 2009 #11

    HallsofIvy

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    Re: Vectors and calculus, lots of help needed (very urgent)

    As a check on the first problem, the area is just half the lateral area of a cylinder of radius 1/2 and of length 1/2. That is one fourth the circumference of a circle of radius 1/2.
     
  13. Nov 2, 2009 #12

    rock.freak667

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    Re: Vectors and calculus, lots of help needed (very urgent)

    Well I sort of obtained the solutions, so I am reading them to understand. I am just confused to one thing.

    For question 2 , I used dA as d(theta)dr

    and in question 5 I used dA as rd(theta)dr

    How do I know when to use what?
     
  14. Nov 2, 2009 #13

    LCKurtz

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    Re: Vectors and calculus, lots of help needed (very urgent)

    It is almost impossible to tell which question you refer to. It might be helpful if you would edit your original post and number the problems sensibly, using the same number if the problem is repeated. You don't have any question 4 or 5 listed.

    The safest thing to do is always use the parametric form of the surface because it builds in any multiple of drd[itex]\theta[/itex]. For example, if the surface is in the xy plane, your usual polar coordinates give:

    [tex]\vec{R}(r,\theta) = \langle r\cos\theta, r\sin\theta, 0\rangle[/tex]

    Now if you use the equation:

    [tex]dS = |\vec{R}_r \times \vec{R}_\theta|drd\theta[/tex]

    you will see that it gives you the usual r dr[itex]\theta[/itex]. Check it out. This is what you are used to substituting when the surface is in the xy plane. But if your surface isn't in the xy plane, like a paraboloid z = r2, your parameterization is not 0 in the third variable:

    [tex]\vec{R}(r,\theta) = \langle r\cos\theta, r\sin\theta, r^2\rangle[/tex]

    Now if you use the formula

    [tex]dS = |\vec{R}_r \times \vec{R}_\theta|drd\theta[/tex]

    you will find you don't just get r dr[itex]\theta[/itex]. Try it.

    The moral of the story is, if there is any doubt, use the formula

    [tex]dS = |\vec{R}_r \times \vec{R}_\theta|drd\theta[/tex]

    because it always gives the correct multiplier.
     
  15. Nov 2, 2009 #14

    rock.freak667

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    Re: Vectors and calculus, lots of help needed (very urgent)


    thank you, I was looking for a simple statement like this rather than my lecturer randomly putting formulas everywhere. Sorry about the bad number though.
     
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