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Using polar coordinates to evaluate iterated integrals: Sketch the region determined by the limits of the given iterated integrals, and then evaluate the integrals.

π/23π/214r3/2drdθ

Short Answer

Expert verified

π/23π/214r3/2drdθ=625π

Step by step solution

01

Draw the region.

The region determined by the limits is shown below,

02

Evaluate the integral

I=π/23π/214r3/2drdθI=π/23π/2r32+132+114dθI=25π/23π/2r5214dθI=25π/23π/2452-152dθI=2×315π/23π/2dθI=6253π2-π2I=625π

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