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  • 5 years ago

a hole of radius ""b" is drilled inside a sphere of radius "a" find the remaing volume

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  1. anonymous
    • 5 years ago
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    So let's do it: 1. Off course we can use double integral, but let's use single integral with defined limits. 2. Integral (f(x), a, b) means definite integral of fuction f in interval [a, b] a) Let's make uppper side of circle which during rottion in OX give us a sphere in functional view. we can define it like this: f(x)=+sqrt(R^2 - x^2) Integral(PI*((f(x)*)^2)dx,-R, +R) give us a volume of a sphere (You can imagine this of make some real mathematics calcultion to prove it. It's integral summ of simple cylinders) b) We want to substract cylinder from this spere. Let this radius be r. If you imagine the picture. You'll see that a point from which starts "cap" can be calculated like this: f(x) = r <=> r^2 = R^2-(x')^2 <=> (x') = +-sqrt(R^2-r^2); the cylinder can be defined like this: we rotate g(x)=r where x is [-sqrt(R^2-r^2), +sqrt(R^2-r^2)] c) So total volume is Integral(PI*((f(x)*)^2)dx,-x', +x') - Integral(PI*((g(x)*)^2)dx,-x', +x') = Volume. Let's clculate this. Pi* Integral( (R^2-x^2) - r^2)dx, -x', +x') = ... Notice: make variable change x=sin(t) * (x'); dx=cos(t)dt*(x') and all be ok. ----------------------------------------------------------------------------------- If you need help visualizing the cylandrical shells method, take a look at http://mathdemos.gcsu.edu/mathdemos/shellmethod/gallery/gallery.html Just imagine that your taking the sphere volume using a bunch of cylandrical shells, but them leaving a bunch of them out for the hole in the middle. So once you have you're integrated function (4Pix^3)/4), use values R (for the radius of the circle) and r (radius of the cylander) like so: (4Pi(R)^3)/4)-( 4Pi(r)^3)/4)

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