Question

9) The region between the graphs of $y = x^2$ and $y = \sqrt{x}$ is rotated about the line $x = -2$. Use the washer method to determine the volume of the solid produced by the rotation.

          9) The region between the graphs of $y = x^2$ and $y = \sqrt{x}$ is rotated about the line $x = -2$. Use the washer method to determine the volume of the solid produced by the rotation.
        
9) The region between the graphs of y = x^2 and y = √(x) is rotated about the line x = -2. Use the washer method to determine the volume of the solid produced by the rotation.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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The region between the graphs of y=x^(2) and y=sqrt(x) is rotated about the line x=-2. Use the washer method to determine the volume of the solid produced by the rotation. 9) The region between the graphs of y=x2 and y=Vx is rotated about the line x=-2.Use the washer method to determine the volume of the solid produced by the rotation.
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Transcript

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00:01 So for this question, we want to find the volume again.
00:05 I'm painting by rotating the region bounded by x equals 2 times the square root of y, x equals 0, and y equals 9, about, rotated about the y -axis.
00:15 So once again, we want to start off by drawing the region.
00:22 So the x equals 0 is just the y -axis line, so we don't really have to care about that.
00:28 The y equals 9, we can just call it like that.
00:34 Now the x equals 2 times the square root of y.
00:39 So typically the square root of y looks like this kind of shape, or this kind of shape.
00:46 But the 2 times the square of y, basically what it does is that it basically makes it wider.
00:53 So it's still the same shape.
00:55 We don't really care about what it specifically looks like because this is just a rough sketch.
01:00 So if we draw across the y equals 9 line, you'll see that this is just a rough sketch.
01:06 This is the region we are rotating about the x -axis, about the y -axis, excuse me.
01:13 So once again, you'll see that we are actually making circles.
01:20 So this brings us to the cross -sectional area function.
01:24 And in this case, it will be in terms of y equals pi r squared, r is equal to two.
01:34 Because in this case, r is equal to the x value...
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