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Problem

The region bounded by the given curves is rotated…

01:41

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Problem 42 Hard Difficulty

The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method.

$ x = (y - 3)^2 $ , $ x = 4 $ ; about $ y = 1 $


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04:28

WZ

Wen Zheng

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 6

Applications of Integration

Section 3

Volumes by Cylindrical Shells

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Applications of Integration

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Video Transcript

we can use any method for this question. I would recommend cylindrical method, which means that we're gonna if axes for and exes y minus three squared, we're gonna be setting the two X is equal to each other to actually solve for y. So this is what I'm doing. Four said equal to y minus three squared. Which means that why is three plus or minus two? Where did the post remind us come from? You may ask, when we take the square root, we have to account for positive and negative values, which means this is one and five. Both of these values matter. In fact, these there are two bounds. Now recall the fact that our is why minus one is given in the problem. H is gonna be negative y squared plus six y minus five. This was received from four minus parentheses. Y minus three squared. So now we have our h. It's time to plug in to the formula to pi. Times are bounds from 1 to 5 are times H or is why minus one. H is negative y squared plus six y minus five d y. Before taking the bird taking the integral. I would highly recommend simple, fine or individual terms. We'll make it a lot easier to read when we take the intro. Now that we've done this, we know we can take the integral Do the power method, which means increased the experiment by one divide by the new exponents. Now that we have this, we know we have the top bound plugged in minus the bottom bound, which is gonna give us to pi. Times 2 50 Sex divide by 12 which some plus two one 28 pi divided by three

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