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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.

0π0RR2r-r3drdθ

Short Answer

Expert verified

The value of integral is0z0πR2r-r3drdθ=πR44

Step by step solution

01

Given information

The expression is I=0π0RR2r-r3drdθ

02

Calculation

The objective of this problem is to use polar coordinates to describe the solid and evaluate the integral expression.

I=0π0RR2r-r3drdθ

Here, r=0,r=R and θ=0,θ=π

Integrate with respect to r first,

I=0πR2r22-r440Rdθxndx=xn+1n+1+C

Substitute the limits

I=0πR42-R44dθ

I=int0πR44dθ

Now integrate with respect toθ

I=R44[θ]0πI=πR44

Thus, the value of integral is

0z0πR2r-r3drdθ=πR44

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