Area by washers and/or cylindrical shells: other than x,y

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When rotating a region around a line other than the x or y-axis, the volume can be calculated using both the washer and shell methods. The example provided involves the function y=x, bounded by y=0 and x=1, rotated around the line x=-1. The volume calculated using the washer method yielded 10π/3, while the shell method produced 5π/3, indicating a discrepancy. The error likely stems from incorrect setup in the integral expressions, particularly in the shell method's integrand. Accurate setup and understanding of the rotation axis are crucial for obtaining consistent results.
tsamocki
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What if you want to rotate around something other than the x/y axis?

For example:

Homework Statement



y=x, y=0, x=1, rotated around the line x=-1

Homework Equations



NumberedEquation1.gif


or

NumberedEquation1.gif


The Attempt at a Solution



V= ⌠(between 0 and 1)π[1+x]^2 dx

= π(1/3(x)^3+x^2+2x),x=0, x=1

=π((1/3)(1)^3+(1)^2+2(1))-0

= 10π/3

Shell method:

V= ⌠(between 0 and 1)2πx(x+1)dx

=2π((1/2)x^2+(1/3)x^3)x=1,x=0

=2π((1/2)+(1/3))-0

=2π(5/6)

=5π/3

Obviously the two answers do not match (i don't even know if i am correct on either); where am i going wrong?

Thanks in advance!
 
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unless I am mistaken, if you're using the first equation you have to do it with respect to y...

and when you did it the second way, the expression inside of the integral is wrong.
 
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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