Results 1 to 12 of 12

Math Help - Finding the volume of this figure

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    242

    Finding the volume of this figure

    I'm trying to find the volume as shown in the picture I've attached. First of all, would I have to set the functions equal to x rather than y since the figure goes up and down and not left to right, and second of all, I know I'd have to subtract two integrals of something since the figure is hollow in the middle, but I don't know which I'd subtract from which.
    Attached Thumbnails Attached Thumbnails Finding the volume of this figure-picture-2.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by fattydq View Post
    I'm trying to find the volume as shown in the picture I've attached. First of all, would I have to set the functions equal to x rather than y since the figure goes up and down and not left to right, and second of all, I know I'd have to subtract two integrals of something since the figure is hollow in the middle, but I don't know which I'd subtract from which.
    Using the shell method, you can integrate with respect to x\colon

    V=2\pi\int_a^bp(x)h(x)\,dx

    =2\pi\int_0^1(1-x)\left(5\sqrt x-5x^3\right)\,dx

    (here p(x) is the distance from the axis of revolution at x, and h(x) is the height of the region at x). This can be easily integrated.

    Alternatively, solve for x in the two equations to get x=y^2/25 and x=y^{1/3}/\sqrt[3]5\text. Then we have

    V=\pi\int_a^b\left([R(y)]^2-[r(y)]^2\right)dy

    =\pi\int_0^5\left[\left(1-\frac{y^2}{25}\right)^2-\left(1-\frac{y^{1/3}}{\sqrt[3]5}\right)^2\right]dy\text.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,889
    Thanks
    683
    Quote Originally Posted by fattydq View Post
    I'm trying to find the volume as shown in the picture I've attached. First of all, would I have to set the functions equal to x rather than y since the figure goes up and down and not left to right, and second of all, I know I'd have to subtract two integrals of something since the figure is hollow in the middle, but I don't know which I'd subtract from which.
    hollow? ... are we seeing the correct problem? the directions say to find the volume formed by rotating the region R_3 about the line AB, which is x = 1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2008
    Posts
    242
    Quote Originally Posted by skeeter View Post
    hollow? ... are we seeing the correct problem? the directions say to find the volume formed by rotating the region R_3 about the line AB, which is x = 1.
    Definitely hollow. Draw it out.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,889
    Thanks
    683
    your definition of hollow is different than mine.

    to me, hollow means an enclosed empty space.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2008
    Posts
    242
    Quote Originally Posted by Reckoner View Post
    Using the shell method, you can integrate with respect to x\colon

    V=2\pi\int_a^bp(x)h(x)\,dx

    =2\pi\int_0^1(1-x)\left(5\sqrt x-5x^3\right)\,dx

    (here p(x) is the distance from the axis of revolution at x, and h(x) is the height of the region at x). This can be easily integrated.

    Alternatively, solve for x in the two equations to get x=y^2/25 and x=y^{1/3}/\sqrt[3]5\text. Then we have

    V=\pi\int_a^b\left([R(y)]^2-[r(y)]^2\right)dy

    =\pi\int_0^5\left[\left(1-\frac{y^2}{25}\right)^2-\left(1-\frac{y^{1/3}}{\sqrt[3]5}\right)^2\right]dy\text.
    And I'm actually supposed to integrate that? Wow...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2008
    Posts
    242
    Quote Originally Posted by skeeter View Post
    your definition of hollow is different than mine.

    to me, hollow means an enclosed empty space.
    Right. Exactly. There's empty space that you don't want the volume of.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by fattydq View Post
    And I'm actually supposed to integrate that? Wow...
    First, that should be easy to integrate--just expand. Second, I gave you an easier way to do it using the shell method.

    Quote Originally Posted by fattydq View Post
    Right. Exactly. There's empty space that you don't want the volume of.
    Yes, but it isn't enclosed. Hence the difference in definition.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Oct 2008
    Posts
    242
    I tried that method. Ended up with 5x^3/2)divided by(3/2)-5x^4/4-2x65/2)+5x^5/5, plugged in 1 5 times on my calculator, multiplied the result by pi every time. same wrong answer every time.

    2pi not pi, just realized that...I got it now thanks
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Oct 2008
    Posts
    242
    where did the 1-x come from in your shell method work. That's the only thing i don't understand.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by fattydq View Post
    where did the 1-x come from in your shell method work. That's the only thing i don't understand.
    1-x is the distance between a representative rectangle and the axis of revolution (i.e., the distance between x and the line x=1). In other words, it is the average radius of the shell corresponding to each particular value of x\text.

    Had you been revolving about the y-axis, that factor would just be x; for revolution about x=2, it would be (2-x), and so on.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Oct 2008
    Posts
    242
    Quote Originally Posted by Reckoner View Post
    1-x is the distance between a representative rectangle and the axis of revolution (i.e., the distance between x and the line x=1). In other words, it is the average radius of the shell corresponding to each particular value of x\text.

    Had you been revolving about the y-axis, that factor would just be x; for revolution about x=2, it would be (2-x), and so on.
    I'm sorry could you put that into easier terms? According to my textbook the formula is the integral from a to be of 2pix times f(x)dx
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Finding the Angle in this Figure (see the figure)
    Posted in the Geometry Forum
    Replies: 1
    Last Post: December 19th 2011, 02:44 PM
  2. Can't figure out volume problem Calc
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 30th 2010, 11:19 AM
  3. Volume figure of revolution
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 21st 2009, 06:40 AM
  4. [SOLVED] Finding the volume of this figure
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 24th 2009, 11:46 AM
  5. Volume of a Spatial Figure
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 6th 2009, 02:03 PM

Search Tags


/mathhelpforum @mathhelpforum