Results 1 to 2 of 2

Math Help - Volumes

  1. #1
    Member
    Joined
    Jan 2011
    Posts
    156

    Volumes

    The base of S is a circular disk (most would call this a circle!) with radius r. Parallel cross sections perpendicular to the base are isosceles triangles with height h and unequal side of the base.
    a) Set up the integral for the volume of S
    b) By interpreting the integral as an area, find the volume of A.

    So, I took some mushrooms and now I think that I have this visualized (more or less). The base of the triangle straddles the x-axis, so its base is 2y and its height is h. If each end of the base touches the circle, then

    y = \sqrt{r^2 - x^2)

    So, A = \frac{1}{2} \cdot 2 yh = h\sqrt{r^2 - h^2}

    And, V = \int_{-r}^r h\sqrt{r^2 - h^2}dx

    Am I right so far? What I don't see is how to integrate this. I can do a trig substitution for the radical, but what do I do with h? I think that this is a clue that I have done something wrong.

    Can anyone help? Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,932
    Thanks
    764

    Re: Volumes

    Quote Originally Posted by joatmon View Post
    The base of S is a circular disk (most would call this a circle!) with radius r. Parallel cross sections perpendicular to the base are isosceles triangles with height h and unequal side of the base.
    a) Set up the integral for the volume of S
    b) By interpreting the integral as an area, find the volume of A.

    So, I took some mushrooms and now I think that I have this visualized (more or less). The base of the triangle straddles the x-axis, so its base is 2y and its height is h. If each end of the base touches the circle, then

    y = \sqrt{r^2 - x^2)

    So, A = \frac{1}{2} \cdot 2 yh = h\sqrt{r^2 - h^2}

    And, V = \int_{-r}^r h\sqrt{r^2 - h^2}dx

    Am I right so far? What I don't see is how to integrate this. I can do a trig substitution for the radical, but what do I do with h? I think that this is a clue that I have done something wrong.

    Can anyone help? Thanks.
    almost correct. should be ...

    V = \int_{-r}^r h\sqrt{r^2 - x^2}dx

    using symmetry and noting that h is a constant ...

    V = 2h \int_0^r \sqrt{r^2 - x^2}dx

    now do your planned trig sub
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Volumes
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 26th 2010, 07:41 PM
  2. Replies: 6
    Last Post: February 11th 2010, 07:08 PM
  3. Volumes
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 18th 2009, 04:35 AM
  4. Volumes
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 2nd 2008, 04:22 PM
  5. Volumes
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 8th 2008, 12:16 PM

Search Tags


/mathhelpforum @mathhelpforum