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Calculus: Early Transcendental Fun...

6th Edition
Ron Larson + 1 other
ISBN: 9781285774770

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BuyFindarrow_forward

Calculus: Early Transcendental Fun...

6th Edition
Ron Larson + 1 other
ISBN: 9781285774770
Textbook Problem

HOW DO YOU SEE IT? The solid is bounded below by the upper nappe of a cone and above by a sphere (see figure). Would it be easier to use cylindrical coordinates or spherical coordinates to find the volume of the solid? Explain.

images

To determine

To Calculate: Whether it would be easier to use cylindrical coordinates or spherical coordinates to calculate volume of the given solid bounded below by the upper nappe of a cone z2=x2+y2 and above by a sphere x2+y2+z2=4 as shown in figure in the question.

Explanation

Given:

The figure:

Refer to figure in question.

The solid is bounded below by the upper nappe of a cone z2=x2+y2 and above by a sphere x2+y2+z2=4.

Formula used:

Volume of a solid in cylindrical coordinates is,

V=QdV=θ1θ2g1(θ)g2(θ)h1(rcosθ, rsinθ)h2(rcosθ,rsinθ)rdzdrdθ

And volume of a solid in spherical coordinates is,

V=QdV=φ1φ2θ1θ2ρ1ρ2ρ2sinφdρdθdφ

Calculation:

It is clear from the figure that z varies from z=x2+y2 (from upper nape of the cone) to z=4x2y2 (surface of the sphere).

Consider the general equation of sphere that is x2+y2+z2=ρ2

Where ρ is the radius of sphere. Therefore, radius of the provided sphere x2+y2+z2=22 is,

ρ=2unit

Now the cone z2=x2+y2 and sphere x2+y2+z2=4 intersects at,

x2+y2+z2=4z2+z2=42z2=4z2=2

Take square root on both the sides,

z=2

It is known that in spherical coordinate z=ρcosϕ.

Where ϕ is the angle made by plane z2=x2+y2 or 2=x2+y2 to make sphere x2+y2+z2=4.

Substitute the value of z and ρ in equation z=ρcosϕ to find the value of ϕ,

z=ρcosϕ2=2cosϕ22=cosϕϕ=arccos(22)

ϕ=π4

Thus, bounds of ϕ is, 0ϕπ4

For sphere angle θ varies from 0θ2π

Hence, volume of provided solid is,

QV=ϕ

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