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Welcome to this lesson.
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In this lesson we will evaluate the double integral.
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So here we have integral from 0 to 1 dr.
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We have integral from 0 to pi on 4 r cos squared theta d theta.
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So let's rewrite it in appropriate way.
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Let's bring the r here.
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We have 0 to 1 dr.
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Then let's have the integral from 0 pi on 4 cos squared theta d theta.
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So here we have cos squared theta which is equals to 1 plus cos 2 theta on 2.
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So making the substitution we would have integral from 0 to 1 r d theta.
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Then this becomes integral from 0 to pi on 4.
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We have 1 plus cos of 2 theta on d theta.
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Alright.
01:16
Which we can now write as integral from 0 to 1 r dr.
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So here we have half then integral from 0 to pi on 4 d theta.
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Then plus we have, we can make it like this.
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So that this is integral from 0 to pi on 4 cos 2 theta d theta.
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Alright.
01:58
So here we'll have r squared on 2 from 1 from 0 to 1.
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Then times it.
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We have half there and this is theta.
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This is from 0 to pi on 4.
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Then plus this half again.
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At this time we have 1 on 2 sin 2 theta.
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Evaluating from 0 to pi on 4.
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So if you put all of them together we have 1 squared on 2 minus 0 squared on 2.
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Then this is multiplying 1 on 2 times pi on 4.
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Alright...