Volume of Solid (Washer Method)

dan38
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Homework Statement


Question is:
FInd the volume for area bound between
y = x ^ (1/3)
x = 4y
About the x -axis

I found the volumes from 0 to 8 using the washer method and then multiplied that by 2 since they intersect at y = 0, -2 and 2 x = 0, 8 and -8
Is that wrong?
Cause my answer was double what it should have been..


Homework Equations





The Attempt at a Solution

 
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dan38 said:

Homework Statement


Question is:
FInd the volume for area bound between
y = x ^ (1/3)
x = 4y
About the x -axis

I found the volumes from 0 to 8 using the washer method and then multiplied that by 2 since they intersect at y = 0, -2 and 2 x = 0, 8 and -8
Is that wrong?
Cause my answer was double what it should have been..
What does your integral look like?
 
In integrated the expression:

x^(2/3) - ( x^2/16)
 
Is there something in the problem that you have overlooked? For example is the region that is rotated around the x-axis supposed to be only in the first quadrant? If so, then when you doubled your answer to account for the part of the region in the third quadrant, that would cause your answer to be too large by a factor of two.

Also, can you show the work you did in integrating?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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