Question

Find the surface area of the solid of revolution generated by revolving about the x-axis the arc of y^2 + 4x = 2lny from y = 1 to y = 3.

Answer 1

answer this please.

Answer 2

\[\int\limits_{}^{}2*\Pi*y*\sqrt{dy^2+dx^2}\]

Answer 3

surface area integral

Answer 4

I'm done with those things, but the part making it difficult for me is the integration part. the sqrt(y^4 + y^2 + 1)

Answer 5

\[\int\limits_{1}^{3}2*\pi*y*\sqrt{1+(dx/dy)^2}dy\]

Answer 6

Yes. That's make me overthink. Can you help me?

Answer 7

@dape this is not a surface area integral.

Answer 8

can you find dx/dy ?
2ydy +4dx=2dy/y
(2y-2/y)dy = -4dx
dx/dy = 1/(2y) - y/2

(dx/dy)^2 = 1/(4y^2) - 1/2 + y^2/4
1 + (dx/dy)^2 = 1/(4y^2) + 0.5 + y^2/4

so we need to integrate
2pi*sqrt(0.25 + 0.5y^2 + y^4/4)
pi * sqrt(1+y^2 + y^4)
as you got @Yttrium

Answer 9

So what shall I do next? :D

Answer 10

@Coolsector that's gonna be one hell of an integral.

Answer 11

well lol what can i do

Answer 12

You are gonna have to use elliptic functions to solve it.

Answer 13

but @dape the integral that you wrote wont give the surface area

Answer 14

yeah i agree @Coolsector

Answer 15

Something seems off, I'm getting a negative answer.

Answer 16

maybe we need another approach

Answer 17

btw, how did you get that kind of high smarscore?

Answer 18

what is it ? lol

Answer 19

@dape , i know that it is tempting to write the differential dx or dy
but we really need to work with ds (the length)
http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx
look at surface area formulas there

Answer 20

the hell!!
i wrote :
so we need to integrate
2pi*sqrt(0.25 + 0.5y^2 + y^4/4)
pi * sqrt(1+y^2 + y^4)
but it should be
2pi*sqrt(0.25 + 0.5y^2 + y^4/4)
pi * sqrt(1+2y^2 + y^4)

Answer 21

so
pi sqrt((y^2+1)^2)
pi(y^2+1) !!!

Answer 22

that is easy to integrate..
just a calculation mistake

Answer 23

so the surface area is
pi * (y^3/3 + y) from 1 to 3
which is
28 and 2/3 times pi

Answer 24

first we need to write the formula, since we have the notation about the x-axis, surface area is: on the board
then we pick the 2nd definition of ds since we are given 1 <= y <= 3 as the bounds.