   Chapter 14.5, Problem 35E

Chapter
Section
Textbook Problem

Product DesignA company produces a spherical object of radius 25 centimeters. A hole of radius 4 centimeters is drilled through the center of the object.(a) Find the volume of the object.(b) Find the outer surface area of the object.

(a)

To determine

To calculate: The volume of the object where the radius of the spherical object is given to be 25 centimetres and it is mentioned that a hole of radius 4 centimetres is drilled through the centre of the same object.

Explanation

Given:

The radius of the given spherical object is 25 centimetres. 4 centimetres is the radius of the hole that is drilled through centre of the object.

Formula Used:

The volume can be found by performing double integration on the surface within the limits.

V=(f(x,y))dA

The equation of sphere can be written as x2+y2+z2=r2.

The equation of circle can be written as x2+y2=r2.

Integration formula to be used is xndx=xn+1n+1.

Calculation:

As given in the question, the object is a sphere of radius 25 centimetres; a hole is drilled through centre of the object and the hole is of radius 4 centimetres.

The region is x2+y2+z2=252; it is a sphere and a hole of radius 4 centimetres is drilled through centre of the object, which can be written as x2+y2+z2=42.

Then,

z=f(x,y)=252x2y2

On doing so, the net region obtained appears in the x-y plane in the first quadrant, which can be described as shown below.

For the region being discussed, the limits for both x and y shall be converted to polar points.

x=rcosθ and y=rsinθ; this implies, {by using sin2x+cos2x=1}

x2+y2=r2(cos2x+sin2x)=r2

Thus, the region being discussed is bounded by 4r25 and 0θπ2.

The equation of the region can be written as:

z=252x2y2=252(x2+y2)=252r2

Substitute the equation obtained above in the formula of volume.

In order to account for all the hemispheres, the surface is multiplied by 8.

V=80π2425252r2rdrdθ

Take u=252r2,du=2rdr and apply the formula xndx=xn+1n+1

(b)

To determine

To calculate: The outer surface area of the given object, which is spherical in shape with a radius of 25 centimetres; it also has a hole of radius 4 centimetres is drilled through centre of it.

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