Method of disk (in respect to y)

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SUMMARY

The discussion focuses on calculating the volume of a solid of revolution formed by revolving the curve y = √x around the x-axis and then around the y-axis. The correct volume for the x-axis rotation is established as 8π using the integral formula. When attempting to calculate the volume for the y-axis rotation, the user incorrectly set the limits of integration and the radius, leading to an incorrect result of 16/5 instead of the expected volume. The correct approach involves using the integral of π*(y^2)^2 from 0 to 2.

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  • Understanding of solid of revolution concepts
  • Familiarity with integral calculus
  • Knowledge of the relationship between x and y in functions
  • Experience with calculating volumes using the disk method
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  • Review the disk method for calculating volumes of solids of revolution
  • Learn about the washer method for cases where the region is bounded
  • Study the implications of changing limits of integration in volume calculations
  • Practice additional problems involving rotation about both axes
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Students studying calculus, particularly those focusing on integral calculus and solid geometry, as well as educators looking for examples of volume calculations using the disk method.

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Homework Statement



Revolve the region under the curve y =√x on the interval [0, 4] about the x-axis and
find the volume of the resulting solid of revolution.

YET, NOW, I want to do in respect to y

Homework Equations


integral formula


The Attempt at a Solution



I did dx and I got 8pi which is corrected.

So I decided to do it with dy.

I looked at the graph and I saw the height would be the same (well because of the problem's nature). so a and b will stay the same

integral of pi*r^2
we know r is y^2 (since y^2 = x)
integral of pi * (y^2)^2 from 1 to 4

but I did not get 8pi
why?

thanks.
 
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The integral you evaluated, gives the volume when the curve is rotated about the y-axis, not the x-axis.
 
The original function was y = sqrt of x
so inverse it, we got x = y^2

you were right about the fact. i think the point of a and b should be 0 to 2, since y^2 = x, so (2^2) = 4

integral of pi * (y^2)^2 from 0 to 2
but i got 16/5 instead
 

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