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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.

02π036r-2r2drdθ

Short Answer

Expert verified

The value of integral is

02π036r-2r2drdθ=18π

Step by step solution

01

Given information

The expression is

I=02π036r-2r2drdθ
02

Calculation

Here, r=0,r=3andθ=0,θ=2π

Integrate with respect torfirst,

I=02π6r22-2r3303dθxndx=xn+1n+1+C

Put the limits

localid="1651390685586" I=02π6(3)22-2(3)33dθI=02π9dθ

Now integrate with respect to θ

I=9[θ]02πI=18π

Thus, the value of integral is

02π036r-2r2drdθ=18π

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