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Math Help - application of definite integral

  1. #1
    Junior Member TheMathTham's Avatar
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    application of definite integral

    "A round hole of radius \sqrt{3} is bored through the center of a sphere of radius 2 feet. Find the volume of the piece cut out."

    I have a vague idea on how to do this. \int_{-\sqrt{3}}^{\sqrt{3}} 2\Pi r^{2}+2\Pi rh. Integral of the surface area. Except, I don't know what to do with the extra variable.
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  2. #2
    Member mathemagister's Avatar
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    Quote Originally Posted by TheMathTham View Post
    "A round hole of radius \sqrt{3} is bored through the center of a sphere of radius 2 feet. Find the volume of the piece cut out."

    I have a vague idea on how to do this. \int_{-\sqrt{3}}^{\sqrt{3}} 2\Pi r^{2}+2\Pi rh. Integral of the surface area. Except, I don't know what to do with the extra variable.
    If you think about the diagram, you'll notice that the h is simply the radius of the sphere, 2 feet. Now, you only have one variable.
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  3. #3
    Junior Member TheMathTham's Avatar
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    Quote Originally Posted by mathemagister View Post
    If you think about the diagram, you'll notice that the h is simply the radius of the sphere, 2 feet. Now, you only have one variable.
    I did that, though I didn't get the right answer, I did notice something else. The cylinder would not have a flat top, it would have a bulbous top on each side of the cylinder because of the sphere. That would mean that the cylinder's height would not be 2, it would be (2-2r). The two bulbous lids would have a height of r. But when I tried to add those in, I get a negative number? The correct answer is \frac{28\Pi }{3}\approx 29.3215. My logic is starting to confuse me...
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    Quote Originally Posted by TheMathTham View Post
    "A round hole of radius \sqrt{3} is bored through the center of a sphere of radius 2 feet. Find the volume of the piece cut out."

    I have a vague idea on how to do this. \int_{-\sqrt{3}}^{\sqrt{3}} 2\Pi r^{2}+2\Pi rh. Integral of the surface area. Except, I don't know what to do with the extra variable.
    If you rotate the region enclosed by the circle x^2+y^2=4 and the y-axis and the line x=\sqrt{3} about the y-axis, the resulting solid would what is bored out of the sphere.

    Using shells, the volume would be

    \int_0^{\sqrt{3}}(2\pi{x})(2\sqrt{4-x^2})\,dx
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  5. #5
    Member mathemagister's Avatar
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    Quote Originally Posted by TheMathTham View Post
    I did that, though I didn't get the right answer, I did notice something else. The cylinder would not have a flat top, it would have a bulbous top on each side of the cylinder because of the sphere. That would mean that the cylinder's height would not be 2, it would be (2-2r). The two bulbous lids would have a height of r. But when I tried to add those in, I get a negative number? The correct answer is \frac{28\Pi }{3}\approx 29.3215. My logic is starting to confuse me...
    Okay, I did some researching on the almighty Internet and found this...

    Draw a 2-dimensional picture of a cross-section through the middle. Look at the right hand side. You have a 2-dimensional object bounded on the right by the curve x^2 + y^2 = 4 and on the left by the line x = \sqrt {3}. The points where these curves intersect are (\sqrt {3},1) and (\sqrt {3}, - 1). We want the volume of the solid of revolution formed by rotating this about the y axis.

    Method 1 (the shell method):


    Slice the 2-dimensional object into vertical strips. Each such vertical strip is located at position x, has width dx, and height 2\sqrt {4 - x^2}. Rotating this strip about the vertical axis gives a thin cylindrical shell. Slice this open and lay it out flat. The result is a rectangular slap of width 2\pi x (the circumference of the shell), height 2\sqrt {4 - x^2}, and thickness dx so dV = 4\pi x\sqrt {4 - x^2}\,dx and

    V = \int_{\sqrt {3}}^24\pi x\sqrt {4 - x^2}\,dx

    Method 2 (the "washer" method):

    Slice the 2-dimensional object into horizontal strips. Each such horizontal strip is located at position y and has width dy. Its left endpoint is at x = \sqrt {3} and its right endpoint is at x = \sqrt {4 - y^2}. Rotating this strip about the vertical axis gives a thin flat "washer" of thickness dy. The inner radius of this (annulus) is \sqrt {3} and the outer radius is \sqrt {4 - y^2}. The volume is

    dV = \pi\left(\left(\sqrt {4 - y^2}\right)^2 - \left(\sqrt {3}\right)^2\right)dy = \pi(1 - y^2)dy


    V = \int_{ - 1}^1 \pi(1 - y^2)dy

    I'm sure you know how to solve either of these integrals.
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  6. #6
    Junior Member TheMathTham's Avatar
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    Oh okay. That makes sense. Ione's equation is correct. I was unfamiliar with the equations that produce semi-circles. I'm confused on why the height is 2\sqrt{4-x^{2}} though. I would have thought that it would be f(x) - g(x), aka \sqrt{4-x^{2}} - \sqrt{3}
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    Quote Originally Posted by TheMathTham View Post
    Oh okay. That makes sense. Ione's equation is correct. I was unfamiliar with the equations that produce semi-circles. I'm confused on why the height is 2\sqrt{4-x^{2}} though. I would have thought that it would be f(x) - g(x), aka \sqrt{4-x^{2}} - \sqrt{3}
    Your slices go from the top part of the circle to the bottom part of the circle so the length of each slice is twice the distance from the x-axis to the top part of the circle.
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  8. #8
    Member mathemagister's Avatar
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    Quote Originally Posted by mathemagister View Post
    Okay, I did some researching on the almighty Internet and found this...

    Draw a 2-dimensional picture of a cross-section through the middle. Look at the right hand side. You have a 2-dimensional object bounded on the right by the curve x^2 + y^2 = 4 and on the left by the line x = \sqrt {3}. The points where these curves intersect are (\sqrt {3},1) and (\sqrt {3}, - 1). We want the volume of the solid of revolution formed by rotating this about the y axis.

    Method 1 (the shell method):


    Slice the 2-dimensional object into vertical strips. Each such vertical strip is located at position x, has width dx, and height 2\sqrt {4 - x^2}. Rotating this strip about the vertical axis gives a thin cylindrical shell. Slice this open and lay it out flat. The result is a rectangular slap of width 2\pi x (the circumference of the shell), height 2\sqrt {4 - x^2}, and thickness dx so dV = 4\pi x\sqrt {4 - x^2}\,dx and

    V = \int_{\sqrt {3}}^24\pi x\sqrt {4 - x^2}\,dx

    Method 2 (the "washer" method):

    Slice the 2-dimensional object into horizontal strips. Each such horizontal strip is located at position y and has width dy. Its left endpoint is at x = \sqrt {3} and its right endpoint is at x = \sqrt {4 - y^2}. Rotating this strip about the vertical axis gives a thin flat "washer" of thickness dy. The inner radius of this (annulus) is \sqrt {3} and the outer radius is \sqrt {4 - y^2}. The volume is

    dV = \pi\left(\left(\sqrt {4 - y^2}\right)^2 - \left(\sqrt {3}\right)^2\right)dy = \pi(1 - y^2)dy


    V = \int_{ - 1}^1 \pi(1 - y^2)dy

    I'm sure you know how to solve either of these integrals.
    Do you mean that mine were wrong? Do you know how it that might be?
    Thanks
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