The area of the surface obtained by revolving the curve x Vy+3 for as-3.) n(y +3) dy 1+ 4y + 12 dy 2nVy+3 dy 2n/4y+ 12 dy a T4y+13 dy

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Chapter6: Topics In Analytic Geometry
Section6.6: Parametric Equations
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The area of the surface obtained by revolving the curve x=/y+3 for a<y<b, about the y-axis, is given by which of the
following integrals?
(a and bare constants, and a>-3.)
A(y +3)° dy
2n, /1+
dy
4y+ 12
a.
2nVy+3 dy
b.
27/4y+ 12 dy
a
I/4y + 13 dy
Transcribed Image Text:The area of the surface obtained by revolving the curve x=/y+3 for a<y<b, about the y-axis, is given by which of the following integrals? (a and bare constants, and a>-3.) A(y +3)° dy 2n, /1+ dy 4y+ 12 a. 2nVy+3 dy b. 27/4y+ 12 dy a I/4y + 13 dy
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