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Math Help - Geometric Applications of Definite Integrals.

  1. #1
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    Geometric Applications of Definite Integrals.

    Volume of solid of revolution the rotation curve:
    x^2 + y^2 = 25 around the y-axis

    Outcome=500*PI/3

    I have to be calculated using definite integral. But I don't really know how to calculate it. Can someone help me please?
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  2. #2
    Junior Member BariMutation's Avatar
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    You'll need a triple integral, and in my opinion, it'd be easiest to do this using spherical coordinates. If any of this is confusing, let me know and I can simplify it using only rectangular coordinates.

    You'll have a triple integral in the order of d(rho)d(phi)d(theta). The limits on theta will be 0 to 2pi (because it's a sphere), the limits on phi will be 0 to pi/2 (the angle it forms with the z axis), and the limits on rho will be 0 to 5 (the distance from the origin to the curve). In the integrand will be 1, since you're finding volume. Understand, or do I need to use rectangular coordinates?
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  3. #3
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    That is, of course, a sphere with radius 5 so you could use the formula V= \frac{4}{3}\pi r^3= \frac{500}{3}\pi as a check.

    There are many ways to integrate this. The simplest is probably to use "disks" perpendicular to the y-axis. Solving x^2+ y^2= 25 for x, x= \sqrt{25- y^2} in the first quadrant so a disk at a specific y would have radius from x= 0 to x= \sqrt{25- y^2}. The area of such a disk is pi r^2=\pi (25- y^2). Imagining it to have thickness \Delta y (since the thickness is measured along the y-axis), its volume would be \pi (25- y^2)\Delta y. The volume of all of those disks is the sum sum \pi (25- y^2)\Delta y which, in the limit as \Delta y goes to 0, becomes the integral \int_{y= -5}^5 \pi (25- y^2) dy.

    Another, slightly harder method would be to use "shells". Imagine the sphere made up of many small cylinders with axis along tye y-axis and radius "x". Each such cylinder would have radius x and height 2y= 2\sqrt{25- x^2) (2 because it goes both up to \sqrt{25- x^2} and down to -\sqrt{25- x^2}. Imagining that flattened out to a rectangle with length 2\pi x, the circumference of the cylinder, and height 2\sqrt{25- x^2}, it would have an area of 4\pi x\sqrt{25- x^2}. Taking the thickness of that cylinder to be \Delta x (since the thickness is now measured along the y-axis), the volume of such a cylinder would be 4\pi x\sqrt{25- x^2}\Delta x and the volume of all the cylinders together would be \sum 4\pi x\sqrt{25- x^2}\Delta x. Taking the limit as \Delta x goes to 0, that becomes the integral \int_{x=0}^5 4\pi x\sqrt{25- x^2}dx.

    Try integrating both ways and see if you don't get 500\pi/3. Do you see why the first integral is from y= -5 to 5 and the second from x= 0 to 5?
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    MHF Contributor Also sprach Zarathustra's Avatar
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    You can use the formula of a ball: V=(4/3) pi*R^3 (Your R is 5) :-)

    V=2\pi\int^5_0 x^2dy=2\pi\int^5_0 (25-y^2)dy=...
    Last edited by Also sprach Zarathustra; December 10th 2010 at 12:22 PM.
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    It has to be calculated in rectangular coordinates.
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    x^2+y^2=25 \Leftrightarrow x^2+y^2=5^2. It's a central circle with radius r=5. Rotating it around the y-axis would give a sphere.

    Volume of the solid of revolution generated with a curve y=f(x), x\in[a,b] revolving around the y-axis can be calculated using the following expression V_y=2\pi\int\limits_{a}^{b}x\cdot f(x) dx

    Rotating only the quarter of a circle (one in the first quadrant)
    y=\sqrt{25-x^2}, x\in[0,5]
    would give you one half of the solid of revolution (and half of the volume needed), therefore you should multiply the result by 2.
    V=2\cdot 2\pi \int\limits_{0}^{5}x\cdot\sqrt{25-x^2}dx
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  7. #7
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    Thank you so much all for your help.
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