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*rewrite the equation as y=\(\sqrt{x+4}\)

|dw:1365190324765:dw|
\[\int 2\pi ~r~h\]

r = c-x, h = sqrt(x), and dx from 0 to c

omg thank you - that applies for shell method as well correct?

that IS the shell method :)

oh shoot - is there any way to do thi with the disk method?

this problem is listed in the disk method section of my textbook, but i'm not sure how it's done...

\[\int~2\pi r\]
\[r=g(y)\text{ ; which is the inverse of f(x)}\]and dy from 0 to f(c)

inverse of y = sqrt(x); x = y^2

r = x = y^2 - c
:)

the only issue that may present itself with that setup tho is that you get a negative summation

why did i do 2pi r? thats circumference :/ .... area is
\[\int pi r^2\]