Application with Shell Integration Method

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The discussion focuses on using the shell method to calculate the volume of a solid formed by rotating the area bounded by y=x², y=0, and x=1 about the x-axis. The initial attempt yielded a volume of ∏/3, which is incorrect; the correct answer is ∏/5. The error arose from incorrectly defining the height in the shell method, specifically using (1-y) instead of (1-√y). To correct the solution, the proper expression for the shell height should be (1-√y). This adjustment will lead to the correct volume calculation.
Saterial
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Homework Statement


Use the shell method to find the volume of the solid obtained by rotating the region bounded by y=x2, y=0 and x=1 about the x-axis.


Homework Equations


lim Ʃ2∏RhΔw
Δw->0


The Attempt at a Solution


I realize this is hard to visualize without a graph. I uploaded how I split the graph.

Solved for y and found y=0, 1.

lim Ʃ2∏RhΔw
Δw->0
=lim Ʃ2∏y(1-y)Δy // made it 1-y because y=x2.
Δy->0
=lim Ʃ2∏(y-y2)Δy
Δy->0
=2∏∫y-y2dy from 0->1
=2∏[y2/2-y3/3] from 0->1
=∏/3

Using Fundamental Theorem of Calculus, volume was found to be ∏/3. However this answer is incorrect. The answer is suppose to be ∏/5. I solved the question initially using the washer method and obtained the answer of ∏/5 but this question asks specifically to use shell method. I don't know how I can fix what I did wrong in this question. I assume I set the wrong h value. I also always get confused on what variable to use x or y, when taking into about if the rotation is about the x or y axis.

Any help would be great thanks!
 

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Saterial said:

Homework Statement


Use the shell method to find the volume of the solid obtained by rotating the region bounded by y=x2, y=0 and x=1 about the x-axis.


Homework Equations


lim Ʃ2∏RhΔw
Δw->0


The Attempt at a Solution


I realize this is hard to visualize without a graph. I uploaded how I split the graph.

Solved for y and found y=0, 1.

lim Ʃ2∏RhΔw
Δw->0
=lim Ʃ2∏y(1-y)Δy // made it 1-y because y=x2.

The quantity in parentheses is xright-xleft. You have (1-y). Now ##x_{right}=1## alright, but since ##y=x^2##, ## x_{left} =\sqrt y## so you should have ##(1-\sqrt y)## there. That will fix it.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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