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Math Help - Integral Volume

  1. #1
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    Integral Volume

    Integral Volume-semi.gifIntegral Volume-ice.gifIntegral Volume-square.gifIntegral Volume-eq.gif
    I have the examples above and we are supposed to get the volume of each based on the cross sections above. The circle is x^2+y^2=16
    Given that Volume = integral between a and b of A(x) dx so I came up with:

    Circle: 4 (integral between -4 and 4) (16-x^2) dx
    Eq Tri: (integral between -4 and 4) (sq rt 3) (16-x^2)
    Semi: (integral between -4 and 4) (pi/2) (16-x^2)
    Isoc: (integral between -4 and 4) (16-x^2)

    But those aren't working out?
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  2. #2
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    Quote Originally Posted by BooGTS View Post
    Click image for larger version. 

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    I have the examples above and we are supposed to get the volume of each based on the cross sections above. The circle is x^2+y^2=16
    Given that Volume = integral between a and b of A(x) dx so I came up with:

    Circle: 4 (integral between -4 and 4) (16-x^2) dx
    Eq Tri: (integral between -4 and 4) (sq rt 3) (16-x^2)
    Semi: (integral between -4 and 4) (pi/2) (16-x^2)
    Isoc: (integral between -4 and 4) (16-x^2)

    But those aren't working out?
    Square.
    Last edited by dwsmith; January 25th 2011 at 10:46 AM.
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  3. #3
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    Where I say "Circle" It should have read square and I apologize for the pictures not loading in the correct order for my equations.
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  4. #4
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    Quote Originally Posted by BooGTS View Post
    Where I say "Circle" It should have read square and I apologize for the pictures not loading in the correct order for my equations.
    So no circle then?
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  5. #5
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    Quote Originally Posted by BooGTS View Post
    Click image for larger version. 

Name:	semi.gif 
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ID:	20594Click image for larger version. 

Name:	ice.gif 
Views:	37 
Size:	4.4 KB 
ID:	20595Click image for larger version. 

Name:	square.gif 
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Size:	8.6 KB 
ID:	20596Click image for larger version. 

Name:	eq.gif 
Views:	41 
Size:	6.5 KB 
ID:	20597
    I have the examples above and we are supposed to get the volume of each based on the cross sections above. The circle is x^2+y^2=16
    Given that Volume = integral between a and b of A(x) dx so I came up with:

    Square: 4 (integral between -4 and 4) (16-x^2) dx
    Eq Tri: (integral between -4 and 4) (sq rt 3) (16-x^2)
    Semi: (integral between -4 and 4) (pi/2) (16-x^2)
    Isoc: (integral between -4 and 4) (16-x^2)

    But those aren't working out?
    Quote Originally Posted by dwsmith View Post
    So no circle then?
    It should read like in the quote above. The circle on the bottom is still x^2 + y^2 = 16, but the cross section is a square, eq triangle, isoc triangle, and semicircle. Sorry for the confusion I created!
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  6. #6
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    Quote Originally Posted by BooGTS View Post
    It should read like in the quote above. The circle on the bottom is still x^2 + y^2 = 16, but the cross section is a square, eq triangle, isoc triangle, and semicircle. Sorry for the confusion I created!
    Area of a squate is A=s^2

    x=\sqrt{16-y^2}

    s=2\sqrt{16-y^2}

    s^2=4(16-y^2)=A(y)

    \displaystyle\int_a^b A(y)dy \ \text{or} \ 2\int_0^b A(y)dy
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  7. #7
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    Quote Originally Posted by BooGTS View Post
    Click image for larger version. 

Name:	semi.gif 
Views:	40 
Size:	5.7 KB 
ID:	20594Click image for larger version. 

Name:	ice.gif 
Views:	37 
Size:	4.4 KB 
ID:	20595Click image for larger version. 

Name:	square.gif 
Views:	44 
Size:	8.6 KB 
ID:	20596Click image for larger version. 

Name:	eq.gif 
Views:	41 
Size:	6.5 KB 
ID:	20597
    I have the examples above and we are supposed to get the volume of each based on the cross sections above. The circle is x^2+y^2=16
    Given that Volume = integral between a and b of A(x) dx so I came up with:
    With x^2+ y^2= 15,  y= \pm\sqrt{16- x^2} so for given x the base in each figure is 2\sqrt{16- x^2}

    Circle (but apparently you meant square): 4 (integral between -4 and 4) (16- x^2)dx
    The area of a square of base s is s^2. Here, s= 2\sqrt{16- x^2} and its square is 4(16- x^2).
    Yes, that looks like exactly what you have.

    Eq Tri: (integral between -4 and 4) (sq rt 3) (16-x^2)
    An equilateral triangle with base s has height \frac{\sqrt{3}}{2}s and so area \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)\left(2\  sqrt{16- x^2}\right)\left(2\sqrt{16- x^2}\right).
    Looks to me like you are missing a factor of 2.

    Semi: (integral between -4 and 4) (pi/2) (16-x^2)
    The area of a semi-circle with diameter s is \frac{\pi}{4}s^2 so with s= 2\sqrt{16- x^2}, the area will be \pi(16- x^2)
    Again, you are a factor of 2 off.

    Isoc: (integral between -4 and 4) (16-x^2)
    The height of an isosceles triangle is independent of the base.
    You can't find the volume without knowing what the height is supposed to be.

    But those aren't working out?
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