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Each of the integrals or integral expressions in Exercises 39-46 represents the volume of a solid in 3. Use polar coordinates to describe the solid, and evaluate the expressions.

h02π0Rr-r2Rdrdθ

Short Answer

Expert verified

The value of integral is

h02π0Rr-r2Rdrdθ=πhR23

Step by step solution

01

Given information

The expression isI=h02π0Rr-r2Rdrdθ

02

Calculation

Here, r=0,r=Rand θ=0,θ=2π

Integrate with respect to r first,

I=h02πr22-r33R0Rdθxndx=xn+1n+1+C

Put the limits

I=h02πR22-R33Rdθ

localid="1651390798747" I=h02πR26dθ

Now integrate with respect to θ

I=hR26[θ]02πI=hR26[2π]

I=πhR23

Thus, the value of integral is

localid="1651390833501" h02π0Rr-r2Rdrdθ=πhR23

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