Calculus describing the solid
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Calculus describing the solid

[From: ] [author: ] [Date: 13-01-23] [Hit: ]
So in this case we have that a=-1, b =1.x^2 = (1-y^2)^2 so x = +-(1 - y^2). Draw this graph to see that its two sideways parabolae facing opposite directions. The region between 1 and -1 describes their enclosed area.Refer to the link in the source for a graph of the situation: the solid formed is the bulb between y=1 and y=-1 rotated around the y axis.......
This integral represents a solid. I am supposed to describe the solid. Please help.

Pi(integral from -1 to 1) (1-(y^2))^2 dy

Of course I am confused I really don't know where to start. Is it sorta like working backwards? I know how to find the shape when I am given formulas but not when I have the integral.

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Yeah it's like working backwards, that's quite confusing actually. But remember that using the method of discs, the volume of a solid rotated around the y axis is:
V = Pi(integral from a to b) x^2 dy
So in this case we have that a=-1, b =1. And also that
x^2 = (1-y^2)^2 so x = +-(1 - y^2). Draw this graph to see that it's two sideways parabolae facing opposite directions. The region between 1 and -1 describes their enclosed area.

Refer to the link in the source for a graph of the situation: the solid formed is the 'bulb' between y=1 and y=-1 rotated around the y axis.
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