Volume sphere calculus

In calculus, an application of integral calculus is the sphere of revolution. One of the definite shapes is sphere. Some of the examples of solids are cone sphere cuboids and cylinder. A sphere of revolution is the process of rotating a plane curve of the sphere on some straight line to get a ball figure. A straight line lies on the same plane of the sphere. Here we are going to study about the sphere. Sphere has points, no straight lines. The straight lines in the plane are shortest between the points. The Volume of a Sphere is equal to the radius of the ball in the geometry. A sphere is totally symmetric around with the center of all points on the surface of equal distance.

A sphere is a perfectly round geometrical object in three-dimensional space, such as the shape of a round ball. Like a circle in three dimensions, a perfect sphere is completely symmetrical around its center, with all points on the surface lying the same distance r from the center point. This distance r is known as the radius of the sphere. The maximum straight distance through the sphere is known as the diameter of the sphere. It passes through the center and is thus twice the radius.



Volume of the sphere (V) = 4/3 * Pi* r3 cubic units.

Surface area of the sphere = 4 * Pi * r2 square units.

The formula to find the radius of a sphere formula is r = root(3)((3 * volume )/( 4* pi))
Formulas on Volume Sphere Calculus:-

Generally the volume of a solid is calculated as the area of the base times its height as long the area is constant throughout the height of the solid. But this concept can not be directly applied to find the volume of a sphere because the area changes with every cross section of the sphere.

Archimedes, the famous mathematician, engineer and scientist around the year 250 B.C framed a formula, which after many centuries was proved to be right.

Sphere

Volume of a sphere is =4/3 pi r3

Where r = radius

pi  = 3.14

Operations of volume sphere calculus:-

The volume of a sphere is (4/3)pir3.

the volume of a sphere is the sum of the circular cross-sections that make it up. Since r is different for each cross-section, you put in the variable x and get: pix2.

The height of the sphere can be represented by the change in y(dy) so now we get the integral:-

int { (pi )x2*dy }

Since we need the variable of integration in terms of y, I went to the equation of a circle.

x2 + y2 = r2

x2 = r2 - y2

Substituting that into the integral we get the final integral:-

int { (pi )(r2-y2)*dy }

Volume of a sphere – by Archimedes formula

Archimedes found after several experiments that the volume of a sphere and also its surface area is exactly (2)/(3) rd of the volume and the surface area of a cylinder with the same outer dimensions.

Volume of a sphere - emphirical

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Let r be the radius of the sphere. Since the over all dimensions of both the sphere and the cylinder are same, the height of the cylinder is 2r.

Under this condition the volume of the cylinder is,

Area of the base x Height of the cylinder.

=  Πr2 x 2r = 2Πr3

Therefore, as per Archimedes formula the volume of the sphere is,

((2)/(3) )( 2Πr3) = ((4)/(3) )Πr3

So much happy about this result by himself, Archimedes wished a cylinder and globe be placed on his tomb! (This wish was fulfilled)

Volume of a sphere – by calculus method:

With the introduction of the concept of integral calculus, the same formula has been proved mathematically.

A sphere is formed by rotating a semicircle around its diameter.

Volume of a sphere -calculas

Let us assumed a semicircle of radius r with center at origin is rotated about  y-axis. In such a case, the equation of the circle is x2 + y2 = r2 or x2 = r2 – y2

The volume of surface of revolution in this case is the volume of the sphere.

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Let us take infinitely  small slice of the semicircle at any point with the radius at that point as x. Being infinitely small, the slice may be considered as a rectangle of height x and width dy. When this rectangle is revolved around y – axis, it generates a cylinder of volume,  dv = Πx2dy

Therefore, the volume generated by the entire circle which is the volume of a sphere is given by,

V = $\pi\int_{-r}^{r}x^2dy$  (since Π is a constant and can be taken out of the integration symbol)

= $\pi\int_{-r}^{r}(r^2 - y^2)dy$

= Π [(r2)(2r) – ((1)/(3) )(2r3) = 2 Πr3 – ((2)/(3) ) Πr3 = ((4)/(3) ) Πr3
Example Problems on Volume Sphere Calculus:-

Example 1:

Calculate the volume of the sphere of revolution produced when the area over y2= x2 up to x-axis surrounded by the co-ordinates at x=1 and x = a is revolved completely about the x-axis.

Solution:-

The formula to find the required volume sphere =int_a^b pi y^2dx

= pi int_a^1x^2dx

=pi [ (a^3)/(3) -(1)/(3) ]

= (pi)/(3) [a3-1]

The volume of the sphere of revolution is =(pi)/(3) [a3-1]

Example 2:

Find the volume produced by revolving the loop of the curve y2= (20-x) completely about the x-axis between O (0, 0) and A (1, 0).

Solution:

The loop can be completely rotate x-axis means the area can be rotated about x-axis between O and A.

Hence, the volume of the sphere of revolution is,

=int_a^b pi y^2dx

= [(20- (1)/(2) )-(0-0)]

=pi ((40-1)/(2) ).

pi ((39)/(2) )

The volume of the sphere of revolution is pi ((39)/(2) ).

Example3:-

Example problem for sphere:

Diameter of the sphere is = 16cm find the volume of the sphere.

Solution:

Given that diameter is = 16cm

Diameter is given so we need to find the radius of the given sphere

Radius = 16 / 2 = 6

Radius = 6

Volume of a sphere is =4/3 pi r3

r – Radius

pi  – 3.14

Substitute the r, pi values in to the formula

=4/3 (3.14) (8)3

=4/3 (3.14) (512)

= 2 143.57333

The volume of a sphere is = 2 143.57333.

Example 4: Calculate the radius of sphere whose volume is 1000cm3.

Solution:

Given that, the volume of a sphere is 1000cm3.

The formula to find the radius of a sphere formula is r = root(3)((3 * volume )/( 4* pi))

Now plug in the value of volume in the formula.

r = root(3)((3 * volume )/( 4* pi))

r = root(3)((3 * 1000 )/( 4* 3.14))

r = root(3)(3000/12.56)

r = root(3)(238.85)

r = 6.2cm.