Results 1 to 9 of 9

Math Help - Volume of solids, cross sections, and fluid force

  1. #1
    Member
    Joined
    Aug 2009
    Posts
    101

    Volume of solids, cross sections, and fluid force

    Alright, so I have my test review in my hand. This semester's instructor doesn't use coursecompass and the demo class we get is useless. He also doesn't answer questions over e-mail. This was fine and all back in the day that there were study groups for calculus up at my school, but there aren't any now. SO! End rant.

    Have answers to the first two, it seems!
    Spoiler:

    The first problem I have a problem with, the answer is 16pi/7 . The problem is:
    1. "Find the volume of the solid formed when the region bounded by the curves y = x^3+1 , x=1, and y=0 is rotated about the X axis".

    My problem is I never really got the concepts of what to do where and when. I graph it 2 dimensionally, I get a shape sort of like a curved rhombus. I get to pick what method I use for this.

    I think, using the disk method, x^3+1 would be the disk radius and the differential volume would be dv=pi*(x^3+1)^2 dx. Am I right so far? Here I get totally lost.

    How would I use the shell method here?


    2. "Find the surface area of a solid formed by rotating the region bounded by x^3/9, the X axis, and 0 <= x <= 2 about the X-axis" , the answer for this is 98pi/81. I'm missing the same concept from the first one.


    3. "The base of a solid is the parabolic region {(x,y)|y^2 <=x<=1} Cross sections perpendicular to the X-axis are squares. Find the volume." The answer is 2.

    I've never seen that syntax before and I have no idea what he's asking.

    4. "Find the total hydraulic force on a dam in the shape of an equilateral triangle with one vertex pointing down, if the side of the triangle is 100 feet and the water is even with the top. delta = 62 lb/ft^3, P= F/A, P = delta*h" Answer is 7,775,000 LB.

    I'll be honest, I do not have enough of a concept of what's going on here. I know we're talking about a triangle with points at x=-100, x=100, and y=-100. I know somewhere in this I use the integral from A to B of w*( strip depth ) * L(y), but I'm sitting here with the verbatim what he worked in class and don't know what's happening.


    Any help is appreciated! Thanks guys!
    Last edited by Wolvenmoon; January 30th 2011 at 04:33 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398
    "8. Do not ask too many questions in one thread. Do not ask more than two questions in a post. "
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,519
    Thanks
    1404
    The shape of a solid of revolution can be thought of as a number of small cylinders being stuck together. So its volume can be approximated by summing the volumes of these cylinders. As you make the height of each cylinder smaller and increase the number of cylinders, the approximation converges on the exact volume.

    If you are rotating about the \displaystyle x axis, the radius of each cylinder at any point \displaystyle x is equal to the corresponding \displaystyle y value, and call the height of this cylinder \displaystyle dx, because the height of this cylinder is a small change in \displaystyle x.

    So the volume of this cylinder is \displaystyle \pi r^2 h = \pi y^2\,dx.

    The volume of your shape is the sum of all these cylinder volumes, so

    \displaystyle V \approx \sum{\pi y^2 \,dx}

    and as you make each cylinder's heights smaller and create an infinite sum of cylinders, the sum converges on an integral and the approximation becomes exact.

    So \displaystyle V = \int_a^b{\pi y^2\,dx}, where \displaystyle a,b are the endpoints of \displaystyle x.


    Do you think you can do Q1 now?
    Last edited by Prove It; January 30th 2011 at 04:19 PM. Reason: Mixed up the axes ><
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398
    Quote Originally Posted by Wolvenmoon View Post
    ...
    The first problem I have a problem with, the answer is 16pi/7 . The problem is:

    1. "Find the volume of the solid formed when the region bounded by the curves y = x^3+1 , x=1, and y=0 is rotated about the X axis".

    My problem is I never really got the concepts of what to do where and when. I graph it 2 dimensionally, I get a shape sort of like a curved rhombus. I get to pick what method I use for this.

    I think, using the disk method, x^3+1 would be the disk radius and the differential volume would be dv=pi*(x^3+1)^2 dx. Am I right so far? Here I get totally lost.

    ...Any help is appreciated! Thanks guys!
    dV=pi*(x^3+1)^2 dx appears to be correct. Note: (x^3+1)^2 = x^6 +2 x^3 +1 .

    What limits of integration should you use?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398
    Quote Originally Posted by Prove It View Post
    The shape of a solid of revolution can be thought of as a number of small cylinders being stuck together. So its volume can be approximated by summing the volumes of these cylinders. As you make the height of each cylinder smaller and increase the number of cylinders, the approximation converges on the exact volume.
    ...
    What Prove It describes here is generally referred to as the disk method for revolving a region about the x-axis !

    A disk of infinitesimal thickness can also be thought of as a cylinder of infinitesimal height, as Prove It refers to it.
    Last edited by SammyS; January 30th 2011 at 04:40 PM. Reason: The original comment no longer applies.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,519
    Thanks
    1404
    Yes, I stupidly mixed up the axes... It's been edited now.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,519
    Thanks
    1404
    Quote Originally Posted by Wolvenmoon View Post
    2. "Find the surface area of a solid formed by rotating the region bounded by x^3/9, the X axis, and 0 <= x <= 2 about the X-axis" , the answer for this is 98pi/81. I'm missing the same concept from the first one.
    To find surface areas of solids of revolution, divide the region into cylinders and sum the curved surface area of each cylinder.

    If you choose two close points \displaystyle x_{i-1} and \displaystyle x_i, then each cylinder will have a radius of \displaystyle y^*, where \displaystyle y^* corresponds to \displaystyle x^* \in (x_{i-1}, x_i).

    Call the height of this cylinder \displaystyle ds, which is the distance between \displaystyle (x_{i-1}, y_{i-1}) and \displaystyle (x_i, y_i).
    By Pythagoras, \displaystyle ds = \sqrt{dx^2 + dy^2}, where \displaystyle dx is the change in \displaystyle x and \displaystyle dy is the change in \displaystyle y.

    Therefore, the curved surface area of each cylinder is \displaystyle 2\pi rh = 2\pi y^*ds

    \displaystyle = 2\pi y^* \sqrt{dx^2 + dy^2}

    \displaystyle = 2\pi y^* \sqrt{dx^2\left(\frac{dx^2 + dy^2}{dx^2}\right)}

    \displaystyle = 2\pi y^* \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx.

    So the surface area of the solid of revolution is

    \displaystyle SA \approx \sum 2\pi y^*\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx

    which, as you make \displaystyle ds small and increase the number of cylinders, this approximation becomes exact and the sum converges on an integral.

    So \displaystyle SA = \int_a^b{2\pi y\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx}, where \displaystyle a,b are the endpoints of \displaystyle x.


    Can you do Q2 now?
    Last edited by Prove It; January 30th 2011 at 08:20 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Aug 2009
    Posts
    101
    Thanks guys!

    Sorry about all the questions at once, I figured that since the first three are probably the same concept I might as well include them.

    I'm going to go after Q1 and Q2 now. ( Editing first post )
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Aug 2009
    Posts
    101
    Got #1. My limits were off. For some reason I went off and took the place where the graph intersects X=1 ( Y=2 ) as my upper limit of integration. Hence my problem.

    Just to overcomplicate it and get the concept, how would I go after it using the shell method? It starts off as The
    integral from 0 to 2 of 2pi*(x^3+1)...then I lose the shell height. ( P.S., mixing these methods is what stubbed me in the toe. )

    On #2...using the definition of the area of a surface, S=Integral of A to B of 2pi*y* sqrt(1+dy/dx)^2), I got

    2pi * the integral from 0 to 2 of x^3/9*sqrt(1+(x^2/3)^2) which is equal to 98pi/81.

    Thanks guys! Do I need to create a new thread for 3 and 4?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Volume by square cross sections
    Posted in the Calculus Forum
    Replies: 0
    Last Post: September 7th 2010, 06:43 PM
  2. Volume of a solid with known cross sections
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 13th 2009, 08:47 PM
  3. Solids of Known Cross sections(volume)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 20th 2009, 04:02 AM
  4. Help with volume of cross-sections please!
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 16th 2009, 11:55 AM
  5. Volume By Cross-Sections Help
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 6th 2008, 03:50 PM

Search Tags


/mathhelpforum @mathhelpforum