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Given the equation of the sphere below:
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| Photo by Math Principles in Everyday Life |
Prove that the volume of sphere is 4/3 π r3
Solution:
From the given figure, let's consider only the 1/8 of the sphere to get the volume and then multiply it by 8 later.
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| Photo by Math Principles in Everyday Life |
Image that there are many tiny cubes that will fill-up the section of a sphere. Let's assume that the dimensions of the cubes, dx, dy, and dz are perfect enough to fill-up the section of a sphere.
dV = dx dy dz
Integrate on both sides of the equation, we have
Next, let's assign the lower and upper limits for each integrals.
Along the x-axis, let's consider the largest cross section of a sphere at xy plane, the cube is moved from 0 to r, and so the lower limit is 0 and the upper limit is r.
Along the y-axis, let's consider the largest cross section of a sphere at xy plane. The cube is moved from 0 to the surface of the cross section. The cross section is a circle with an equation: x2 + y2 = r2 where z2 = 0. Solve for the value of y, we have
Therefore, the lower limit is 0 and the upper limit is
Along the z-axis, let's consider the largest cross section of a sphere at xy plane. The cube is moved from the cross section of a sphere at xy plane to the surface of a sphere. From the equation of the sphere: x2 + y2 + z2 = r2, solve for the value of z, we have
Therefore, the lower limit is 0 and the upper limit is
The volume of a sphere can be written as
The above equation can also be written as
Integrate first with respect to z, consider x and y are the constants
let a2 = r2- x2
and u2 = y2
Using the integration formula,
Integrate with respect to y, consider x as a constant. Therefore,
Finally, integrate with respect to x, we have
Therefore, the volume of a sphere is
