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Find the volume generated by revolving the regions bounded by the given curves about the $x$ -axis. Use the indicated method in each case.$$y=3 \sqrt{x}, y=0, x=4 \quad(\text { disks })$$
Calculus 2 / BC
Chapter 26
Applications of Integration
Section 3
Volumes by Integration
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Lectures
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In mathematics, integratio…
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Let R be the region bounde…
01:08
Let R be the region boun…
04:33
Find the volumes of the so…
01:05
02:18
02:45
02:13
Use the shell method to fi…
01:14
Find the volume of the sol…
04:27
01:46
Use the method of cylindri…
thing with this problem is that we have the square root function but it is tripled. So it is a A vertical stretch by a factor of three. But then we also have the lines, Y equals zero, Which is why is zero then x must equal zero. Uh You know, three times zero is zero. I hope that makes sense because they also give you the bound Of X. equals four. And so as I do the disk method, as they tell me to do the disk method, uh it's pi times the function squared. Uh in terms of X because we're revolving around the X axis. So that's why my bounds are zero and four is a as I mentioned, my ex zero but also they gave us the bound of four and then it's a function that they gave us. Now if it's helpful out encourage students, especially if you're afraid of making a mistake of actually squaring it. So three times serious nine and road X times X is just X key X. So at this point you can evaluate the integral by adding one to the exponent and divide by your new exponent And do that from 0 to 4. And uh I don't know, I think students should be able to do four squared in their head of 16 divided by two is eight times nine is 72. When you subtract 00 squared? Still zero times nine half. Still 0. 72 minus zero is 72 times pi 72. Pi
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