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Math Help - integration, area and volume

  1. #1
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    integration, area and volume

    a) Find the area enclosed between y=1/sqrt(1-x^2), the x axis and the lines x=0 and x=1/2.

    integrating y I got [sin^-1(x)] between limits 1/2 and 0 which equals
    =[sin^-1(1/2)]-[sin^-1(0)]
    =pi/6 units^2

    b) This area is rotated about the y-axis. Find the exact volume of the soild formed.

    This is what I am thinking...........use V= pi integral x^2 dy with limits pi/6 and 0. I rearranged y from part a) to make x^2 the subject and got x^2=1-y^-2.

    Anywho I integrated and got V=5pi/4 units^3

    The answer is pi/6(7sqrt(3)-12)units^3

    I am way off, can someone please show me where I'm going wrong.
    Thank you
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by slaypullingcat View Post
    a) Find the area enclosed between y=1/sqrt(1-x^2), the x axis and the lines x=0 and x=1/2.

    integrating y I got [sin^-1(x)] between limits 1/2 and 0 which equals
    =[sin^-1(1/2)]-[sin^-1(0)]
    =pi/6 units^2

    b) This area is rotated about the y-axis. Find the exact volume of the soild formed.

    This is what I am thinking...........use V= pi integral x^2 dy with limits pi/6 and 0. I rearranged y from part a) to make x^2 the subject and got x^2=1-y^-2.

    Anywho I integrated and got V=5pi/4 units^3

    The answer is pi/6(7sqrt(3)-12)units^3

    I am way off, can someone please show me where I'm going wrong.
    Thank you
    Hmmm... The derivative of \arcsin{x} is \frac{1}{\sqrt{1-x^2}}. So..... the integral of \frac{1}{\sqrt{1-x^2}} is....
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by slaypullingcat View Post
    a) Find the area enclosed between y=1/sqrt(1-x^2), the x axis and the lines x=0 and x=1/2.

    integrating y I got [sin^-1(x)] between limits 1/2 and 0 which equals
    =[sin^-1(1/2)]-[sin^-1(0)]
    =pi/6 units^2

    b) This area is rotated about the y-axis. Find the exact volume of the soild formed.

    This is what I am thinking...........use V= pi integral x^2 dy with limits pi/6 and 0. I rearranged y from part a) to make x^2 the subject and got x^2=1-y^-2.

    Anywho I integrated and got V=5pi/4 units^3

    The answer is pi/6(7sqrt(3)-12)units^3

    I am way off, can someone please show me where I'm going wrong.
    Thank you
    a) is correct.

    Now for b), I would suggest the method of cylindrical shells.

    V=2\pi\int_0^{\frac{1}{2}}\frac{x}{\sqrt{1-x^2}}\,dx.

    However, I get a different solution that the one you claim it to be...
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by slaypullingcat View Post
    [sin^-1(x)]
    Latex, latex, latex, my friend. I had trouble with your association here. If you wrap math tags ( the sigma button ) around your expression there, you'd see the trouble.
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  5. #5
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    Quote Originally Posted by Chris L T521 View Post
    a) is correct.

    Now for b), I would suggest the method of cylindrical shells.

    V=2\pi\int_0^{\frac{1}{2}}\frac{x}{\sqrt{1-x^2}}\,dx.

    However, I get a different solution that the one you claim it to be...
    that answer is coming from the textbook, I just double checked it. Thanks for your post, its a bit over my head but I appreciate your time.
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  6. #6
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    Quote Originally Posted by VonNemo19 View Post
    Latex, latex, latex, my friend. I had trouble with your association here. If you wrap math tags ( the sigma button ) around your expression there, you'd see the trouble.
    Sorry, I know I should make the effrot. I'm two weeks out from my final exam and I'm putting all my time into study.
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by slaypullingcat View Post
    that answer is coming from the textbook, I just double checked it. Thanks for your post, its a bit over my head but I appreciate your time.
    Its hard to get used to all the different techniques.

    Basically, it comes from the definition of the volume of a cylinder: V=2\pi r h.

    In our case, h(x) (or h(y)) is the height of each shell (its usually the function you're considering rotating about a certain axis or line. In our case, its \frac{1}{\sqrt{1-x^2}}). r(x) (or r(y)) is the radius "distance" from the axis of rotation.

    Since we want to revolve around the y axis, we measure the radius of each shell from 0 to 1/2 (in otherwords, r(x)=x for x\in\left[0,1/2\right].

    Now when we put this all together, we have 2\pi\int_0^{\frac{1}{2}}r(x)h(x)\,dx=2\pi\int_0^{\  frac{1}{2}}x\cdot\frac{1}{\sqrt{1-x^2}}\,dx, which is the integral I gave you.

    I hope this clarifies how I came up with that integral.

    See this pdf for more of an explanation on cylindrical shells.
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  8. #8
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    Quote Originally Posted by Chris L T521 View Post
    Its hard to get used to all the different techniques.

    Basically, it comes from the definition of the volume of a cylinder: V=2\pi r h.

    In our case, h(x) (or h(y)) is the height of each shell (its usually the function you're considering rotating about a certain axis or line. In our case, its \frac{1}{\sqrt{1-x^2}}). r(x) (or r(y)) is the radius "distance" from the axis of rotation.

    Since we want to revolve around the y axis, we measure the radius of each shell from 0 to 1/2 (in otherwords, r(x)=x for x\in\left[0,1/2\right].

    Now when we put this all together, we have 2\pi\int_0^{\frac{1}{2}}r(x)h(x)\,dx=2\pi\int_0^{\  frac{1}{2}}x\cdot\frac{1}{\sqrt{1-x^2}}\,dx, which is the integral I gave you.

    I hope this clarifies how I came up with that integral.



    See this pdf for more of an explanation on cylindrical shells.
    Thank you very much. I get it now. Thank you!
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