- #1
Swallow
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Hello there.
Suppose I have a function:
y=3x[tex]^{2/3}[/tex]-1
I want to find the surface area of the solid formed when the part of the curve between x=0 and x=8 is revolved about the x-axis.The curve crosses the x-axis at a point (1/3)[tex]^{3/2}[/tex]
The derivative of the function is:
y'=2x[tex]^{-1/3}[/tex] which is undefined at x=0, so can i still find the surface area by using the limits 0 to 8 in the formula for surface area i.e.
S.A.= Integral of( 2*pi* f(x)*[tex]\sqrt{1+f'(x)^{2}}[/tex]) (with limits 0 to 8).
moreover, if I can find the surface area do i have to break the integral at the point at which the function changes form negative to positive?
Suppose I have a function:
y=3x[tex]^{2/3}[/tex]-1
I want to find the surface area of the solid formed when the part of the curve between x=0 and x=8 is revolved about the x-axis.The curve crosses the x-axis at a point (1/3)[tex]^{3/2}[/tex]
The derivative of the function is:
y'=2x[tex]^{-1/3}[/tex] which is undefined at x=0, so can i still find the surface area by using the limits 0 to 8 in the formula for surface area i.e.
S.A.= Integral of( 2*pi* f(x)*[tex]\sqrt{1+f'(x)^{2}}[/tex]) (with limits 0 to 8).
moreover, if I can find the surface area do i have to break the integral at the point at which the function changes form negative to positive?
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