Results 1 to 5 of 5

Math Help - Surface Area

  1. #1
    Member
    Joined
    May 2006
    Posts
    148
    Thanks
    1

    Surface Area

    Suppose we have a solid S that is bounded by the following surfaces:

    z = 5x^2 + 5y^2 and z = 28 - 2x^2 - 2y^2

    1.) Determine the surface area of S.

    2.) Determine the avg. distance from the z-axis to a point in S.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2006
    Posts
    148
    Thanks
    1
    Could someone please check my work!

    5x^2 + 5y^2 = 28 - 2x^2 - 2y^2

    7x^2 + 7y^2 = 28
    y = +/- sqrt(4-x^2) . . . z = 20

    z = 5x^2 + 5y^2

    f_x = 10x = 10*r*cos(theta)
    f_y = 10y = 10*r*sin(theta)

    z = 28 - 2x^2 - 2y^2
    f_x = -4x = -4*r*cos(theta)
    f_y = -4y = -4*r*sin(theta)

    SA = int(int_R(sqrt(f_(x)^2 = f_(y)^2 + 1)dA

    int(int(sqrt(16r^2*cos^2(theta) + 16r^2*sin^2(theta) + 1)r dr d(theta)..

    Note the limits of integration are, respectively, 0 to 2Pi, 0 to 2..

    I did it out and got 172.6 ...

    PLEASE would someone help me determine if this is right.

    For the avg. distance I'm not sure.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by fifthrapiers View Post
    Suppose we have a solid S that is bounded by the following surfaces:

    z = 5x^2 + 5y^2 and z = 28 - 2x^2 - 2y^2

    1.) Determine the surface area of S.
    .
    The second curve is parabolid which opens down. And First curve is paraboloid which open up. Their intersection forms a surface. Now the second one is one top and first one is below.

    And how do they intersect? Note:
    z=5x^2+5y^2
    z=28-2x^2-2y^2
    Thus,
    5x^2+5y^2 = 28 - 2x^2 - 2y^2
    Thus,
    7x^2+7y^2 = 28
    Thus,
    x^2 +y^2 = 4
    That is they intersect in a circle centered at origin or radius 2.

    Now follow attachment.
    Attached Thumbnails Attached Thumbnails Surface Area-picture2.gif  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2006
    Posts
    148
    Thanks
    1
    Quote Originally Posted by ThePerfectHacker View Post
    The second curve is parabolid which opens down. And First curve is paraboloid which open up. Their intersection forms a surface. Now the second one is one top and first one is below.

    And how do they intersect? Note:
    z=5x^2+5y^2
    z=28-2x^2-2y^2
    Thus,
    5x^2+5y^2 = 28 - 2x^2 - 2y^2
    Thus,
    7x^2+7y^2 = 28
    Thus,
    x^2 +y^2 = 4
    That is they intersect in a circle centered at origin or radius 2.

    Now follow attachment.
    Why do you have 3 integrals?

    I thought SA is given by 2 integrals.

    SA = int(int_R(sqrt(f_(x)^2 + f_(y)^2) + 1)dA

    Am I wrong?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by fifthrapiers View Post
    Why do you have 3 integrals?

    I thought SA is given by 2 integrals.

    SA = int(int_R(sqrt(f_(x)^2 + f_(y)^2) + 1)dA

    Am I wrong?
    No, I looked you were saying to find the volume.

    Instead do this as two seperate problems. First do the upper surface and then do the lower seurface. I worked out their intersections in my fist post.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surface Area of Surface
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 1st 2010, 09:53 AM
  2. Calculate the surface area of the surface
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 26th 2009, 04:03 AM
  3. Lateral Area and Total Surface Area
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 25th 2009, 04:28 PM
  4. Help finding surface area of a surface
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 3rd 2008, 04:11 PM
  5. Volume, Surface Area, and Lateral Surface Area
    Posted in the Geometry Forum
    Replies: 1
    Last Post: April 14th 2008, 11:40 PM

Search Tags


/mathhelpforum @mathhelpforum