Hello temaire Originally Posted by

**temaire** Find the area of the surface generated by revolving the curve $\displaystyle y = \sqrt{x + 1} , 1 \leq x \leq 5$ aroung the x-axis.

$\displaystyle S = \int^{b}_{a} 2\pi y \sqrt{1 + (dy/dx)^{2}} dx$

$\displaystyle S = \int^{5}_{1} 2\pi \sqrt{x + 1} \sqrt{1 + (1/2\sqrt{x + 1})^{2}} dx$

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I used the substitution rule

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$\displaystyle S = 49\pi /3$

Am I correct?

Thanks.

That's exactly what I got. Good job!

Incidentally, you can use a little trick to avoid some of the fractional powers, and make the working a bit easier: $\displaystyle y=\sqrt{x+1}$

$\displaystyle \Rightarrow y^2 = x+1$

$\displaystyle \Rightarrow 2y\frac{dy}{dx}=1$

$\displaystyle \Rightarrow \left(y\frac{dy}{dx}\right)^2 = \tfrac14$

Now write$\displaystyle S = \int^{b}_{a} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^{2}} \;dx$

as:$\displaystyle S = 2\pi\int^{b}_{a} \sqrt{y^2 + \left(y\frac{dy}{dx}\right)^{2}} \;dx$$\displaystyle =2\pi\int_1^5\sqrt{x+\tfrac54}\;dx$

$\displaystyle = ...$ etc

Grandad