Results 1 to 4 of 4

Math Help - Voulme by integration

  1. #1
    Newbie Exotique's Avatar
    Joined
    Apr 2010
    Posts
    18

    Voulme by integration

    A glass vase has the shape of the solid obtained by rotating about the y-axis the area in the first quadrant lying over the x-interval [0,a] and under the graph of y=\sqrt{x}

    Determine how much glass is contained in the vase.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Exotique View Post
    A glass vase has the shape of the solid obtained by rotating about the y-axis the area in the first quadrant lying over the x-interval [0,a] and under the graph of y=\sqrt{x}

    Determine how much glass is contained in the vase.
    Hi Exotique,

    y=\sqrt{x}\ \Rightarrow\ x=y^2

    As the curve is being rotated about the y-axis, we need the limits on the y-axis.
    Also, we integrate the discs with radius=x.

    x=0,\ y=0

    x=a,\ y=\sqrt{a}

    The volume of the vase is \int_{y=0}^{\sqrt{a}}{\pi}x^2dy

    ={\pi}\int_{0}^{\sqrt{a}}y^4dy

    If the glass is the solid part under the graph above the x-axis, with a hollow interior above the graph,

    subtract this integral from {\pi}a^2(\sqrt{a})
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie Exotique's Avatar
    Joined
    Apr 2010
    Posts
    18
    The value I obtained was
    \pi\sqrt{a}^5-\frac{1}{5}\pi\sqrt{a}^5
    =\frac{4\pi}{5}\cdot\sqrt{a}^5

    Is this correct?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Exotique View Post
    The value I obtained was
    \pi\sqrt{a}^5-\frac{1}{5}\pi\sqrt{a}^5
    =\frac{4\pi}{5}\cdot\sqrt{a}^5

    Is this correct?
    Yes,

    if the vase really is that shape.

    We get the same answer by integrating the surface areas of the washers
    (denoted by the 2 red discs).

    This is equivalent to subtracting the interior volume (of air)
    from the volume of a cylinder.

    \int_{0}^{\sqrt{a}}\left({\pi}a^2-{\pi}x^2\right)dy={\pi}\int_{0}^{\sqrt{a}}\left(a^  2-x^2\right)dy

    ={\pi}\int_{0}^{\sqrt{a}}\left(a^2-y^4\right)dy={\pi}(\sqrt{a})^5-{\pi}\frac{(\sqrt{a})^5}{5}
    Attached Thumbnails Attached Thumbnails Voulme by integration-vase.jpg  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 01:54 AM
  2. Replies: 2
    Last Post: November 2nd 2010, 05:57 AM
  3. Replies: 8
    Last Post: September 2nd 2010, 01:27 PM
  4. Replies: 2
    Last Post: February 19th 2010, 11:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 07:58 AM

Search Tags


/mathhelpforum @mathhelpforum