Evaluate surface area of the curve y=sqrt(x+1) from x=1 to x=5 when it is revolved around the x axis

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Evaluate surface area of the curve y=sqrt(x+1) from x=1 to x=5 when it is revolved around the x axis

Mathematics
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Oh boy, surface area. Okay, start with the general equation for surface area: \[SA = \int\limits_{a}^{b}2\pi f(x)\sqrt{1+(f \prime(x))^2}dx\]f(x) is given to you as \[y = \sqrt{x + 1}\]with limits a = 1 and b = 5. The next part of this equation comes from the equation for distance. Take the first derivative of your f(x), square it, add 1, and take the square root. \[f(x) = \sqrt{x + 1}\]\[f \prime(x) = 1/(2 \sqrt{x+1})\]\[\sqrt{1 + (1/(2 \sqrt{x+1}))^2}\]Yeah, this is looks wonderful. It simplifies well though.\[\sqrt{1 + 1/(4 + 4x)}\]\[\sqrt{(4x+5)/(4(x+1))}\]\[1/2\sqrt{(4x + 5)/(x+1)}\]Much better.....kinda. From here you plug this in and simplify more before taking the integral. Do you want the steps for those too?

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