How do you find the volume of a solid of revolution using integration?

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SUMMARY

The discussion focuses on calculating the volume of a solid of revolution defined by the curve x=2y², bounded by y=±6, when rotated around the y-axis. The correct integral setup for this problem is ∫ from -6 to 6 of 2πy(2y²) dy, which represents the volume as the area of circular cross-sections multiplied by the differential width dy. Participants clarify that the symbols in the integral represent multiplication, and emphasize the importance of understanding the radius of the circles formed during rotation.

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Homework Statement



x=2y^2
x=0
y=+-6
rotated around y

Homework Equations


integral 2pi*x(f(x)dx

The Attempt at a Solution



integral from -6 to 6 of 2pi*y*2y^2 dy
but i get something far less than the correct answer
 
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What are all the * symbols supposed to mean? Multiplication?

Just think of this as Area times length. Since it's rotation around the y-axis you can see that it's easiest to differentiate with respect to y. It's going to make a bunch of little circles. What is the radius of those? (your function?).

From there, you need to plug it into the area of a circle to get pi(radius)^2
So if the volume of the solid is area times width, and in the integration dy is the width, what integral would solve this problem?
 

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