Volume of Solid Revolving Region Bounded by x=y^2, x=4 About x=5

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SUMMARY

The volume of the solid generated by revolving the region bounded by the curves x = y² and x = 4 about the line x = 5 can be calculated using the washer method. The correct setup involves defining the outer radius R(y) as y² and the inner radius r(y) as 1. The integral should be evaluated from y = 0 to y = 2, leading to the expression ∏∫(y²)² - (1)² dy. The initial calculation of 16.96 is incorrect, and the discrepancy arises from misapplying the washer method's parameters.

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Homework Statement

Find volume of the solid generated by revolving the region bounded by x = y^2, x=4
about the line x = 5

Homework Equations



the washer method from c to d ∏∫ R(y)2 - r(y)2

The Attempt at a Solution



I set r(y) = 1 and R(y)= y^2

and got the integral from 0 to 2 of ∏∫(y^2)^2-(1)^2

I got an answer of 16.96 but i know this is wrong because the back of the book gives a different answer. Please tell me where I went wrong.
 
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Blonde1551 said:

Homework Statement




Find volume of the solid generated by revolving the region bounded by x = y^2, x=4
about the line x = 5

Homework Equations



the washer method from c to d ∏∫ R(y)2 - r(y)2

The Attempt at a Solution



I set r(y) = 1 and R(y)= y^2

and got the integral from 0 to 2 of ∏∫(y^2)^2-(1)^2

I got an answer of 16.96 but i know this is wrong because the back of the book gives a different answer. Please tell me where I went wrong.
Think about what the washer method involves -- why it's called the "washer" method.

What do r(y) and R(y) represent?
 

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