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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621

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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621
Textbook Problem

A paper drinking cup filled with water has the shape of a cone with height h and semi vertical angle θ . (See the figure.) A ball is placed carefully in the cup, thereby displacing some of the water and making it overflow. What is the radius of the ball that causes the greatest volume of water to spill out of the cup?

Chapter 5.P, Problem 10P, A paper drinking cup filled with water has the shape of a cone with height h and semi vertical angle

To determine

To find:

Maximum radius of a ball that causes the greatest volume of water to spill out of the cup

Explanation

1) Concept:

Let S be a solid that lies between y=a  and y=b. If the cross sectional area of S in the plane Py through y and perpendicular to the y axis is A(y), where A  is a continuous function, then volume of S is

V=abAydy

If the cross section formed is the disk, then find the radius of the disk in terms of y and use

Ay=πradius2

2) Calculation:

Consider a cone whose tip is at the origin, given height h. Assume that the sphere of radius r is inside the cone at a distance of X from the origin.

Volume of sphere inside the cone can be obtained by revolving the portion circle from X-r to h about x-axis.

sinθ=rX.

The equation of the circle having the center at (X, 0) is

x-X2+y2=r2

So, y=r2-x-X2

By using themethod of washer,

Ax=πr2-x-X22

So, V=X-rhπr2-x-X22dx

V=πX-rh[r2-x-X2]dx

V=πr2x-x-X33X-rh

V=πr2h-h-X33-r2(X-r)-X-r-X33

V=πr2h-h-rsinθ33-r2rsinθ-r+r33

V=πr2h-h-rsinθ33-r31sinθ-1+r33

V=πr2h-h-rcosec θ33-r3cosec θ-1+r33

V=πr2h-h-rcosec θ33-r3cosec θ-2r33

V=πr2h-h-rcosec θ33-r3cosec θ+2r33

Now, differentiate with respect to r.

dVdr=π2rh+cosecθ·h-r cosecθ2-3r2cosecθ-2r2

dVdr=π2rh+cosecθ·h-r cosecθ2-3r2cosecθ+2r2

dVdr=π2rh+cosecθ·h2-2hrcosec θ+r2cosec2θ-3r2cosecθ+2r2

dVdr=π2rh+h2cosec θ-2hrcosec2 θ+r2cosec3θ-3r2cosecθ+2r2

Substitute dVdr=0.

π2rh+h2cosec θ-2hrcosec2 θ+r2cosec3θ-3r2cosecθ+2r2=0

2rh+h2cosec θ-2hrcosec2 θ+r2cosec3θ-3r2cosecθ+2r2=0

Convert the above equation into ax2+bx+c=0 form

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