Surface area of rotating curve about $x$axis: $y=(x+1)^{1/2}$
You may write $$ \begin{align} \int_0^3(x+1)^{1/2}\sqrt{1+\frac14(x+1)^{1}}\:dx&=\int_0^3(x+1)^{1/2}\sqrt{1+\frac1{4(x+1)}}\:dx \\\\&=\int_0^3(x+1)^{1/2}\sqrt{\frac{4(x+1)+1}{4(x+1)}}\:dx \\\\&=\int_0^3(x+1)^{1/2}\frac{\sqrt{4(x+1)+1}}{\sqrt{4(x+1)}}\:dx \\\\&=\int_0^3\color{red}{(x+1)^{1/2}}\frac{\sqrt{4x+4+1}}{\sqrt{4}\color{red}{\sqrt{x+1}}}\:dx \\\\&=\frac12\int_0^3\sqrt{4x+5}\:dx \end{align} $$ then you are led to evaluate $$ \begin{align} \int_0^3\sqrt{4x+5}\:dx&=\int_0^3(4x+5)^{1/2}\:dx \\\\&=\left[\frac{1}{4(1/2+1)} (4x+5)^{1+1/2}\right]_0^3 \\\\&=\left[\frac{1}{6} (4 x+5)^{3/2}\right]_0^3 \\\\&=\frac{17\sqrt{17}}6\frac{5\sqrt{5}}6. \end{align} $$
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Landon Tholen
Updated on August 04, 2020Comments

Landon Tholen over 3 years
I'm working on finding the surface area of a revolution of a line about the xaxis. I'm looking for some assistance in determining the best way to solve problems like this.
$y^2 = x+1, \quad 0 \le x \le 3$
I understand the calculus principal behind this method however; I keep getting hung up on what to do with the portion under the radical. So using the surface area formula I'll end up with this:
$$\int_0^3 2\pi (x+1)^{1/2}\sqrt((1+((1/2)(x+1)^{1/2})^2))$$
For this section:
$\sqrt((1+((1/2)(x+1)^{1/2})^2))$
Do I multiply it out to be $\sqrt(1 + ((1/4)(x+1)^{1})$ and go from there? or is there some other way I can do this? I can't see any way to get to usubstitution or solve the radical... Any pointers would be appreciated! Thanks!

Landon Tholen over 7 yearsHow do you get from $$ \int_0^3(x+1)^{1/2}\sqrt{1+\frac14(x+1)^{1}}\:dx $$ to $$ \int_0^3\sqrt{4x+5} $$ ? That's the main part that I'm having trouble seeing the transition for. Edit: Formatting

Olivier Oloa over 7 years@LandonTholen I've edited my answer consequently. Hoping it is Ok now.

Landon Tholen over 7 yearsYes! That makes much more sense now! I was attempting to multiply out the portion under the radical (which does not end well) but I understand how you separate everything to cancel items out. Thanks for the help!!