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The curve $y=\sqrt{x^2+1}, 0\leqslant{x}\leqslant{\sqrt{2}}$, which is part of the upper branch of the hyperbola $y^2-x^2=1$, is revolved about x-axis to generate a surface. Find the area of the surface.

My plan is first to calculate the formula of the surface, then use surface integral to calculate its surface area. Then what is the formula of this surface? Thanks.

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In the plane, $y$ is the distance of the point $(x, y)$ from the $x$, axis. When you revolve the curve about the $x$-axis, this distance should remain the same for every point $(x, y, z)$ on the surface. That is, the distance of $(x, y, z)$ from the $x$-axis should be equal to the distance of the generating point $(x, y)$ from the $x$-axis.

Originally, $y^2 = x^2 + 1$, $0 \le x \le \sqrt 2$. Now, therefore:

$$\boxed{y^2 + z^2 = x^2 + 1,\ 0 \le x \le \sqrt 2}$$

Another way to think of it is that each point $(x, y)$ on the curve generates a circle on the surface with radius equal to the height $y$ of the point.

I hope that you know of formulas that allow you to directly calculate the surface area (and volume) of revolution, rather than finding the equation of the surface and then using integration on that.

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There is a standard formula for area of a surface of revolution obtained by rotating $y=f(x)$ about the $x$-axis, from $x=a$ to $x=b$. It says that area is $$\int_a^b 2\pi f(x)\,ds,$$ where $ds=\sqrt{1+(f'(x))^2}\,dx$.

In our case, $f(x)=\sqrt{x^2+1}$ and therefore $f'(x)=\frac{x}{\sqrt{x^2+1}}$.

Remarks: $1.$ The idea behind the formula is that we look at the little bit of area swept out by the part of the curve from $x$ to $x+dx$. The arclength of this little bit of curve is approximately $ds=\sqrt{1+(f'(x))^2}\,dx$. So this part of the curve sweeps out a "ribbon" with radius $y=f(x)$ and width $ds$. The approximate area of the ribbon is then $2\pi f(x)\,ds$. We "add up" these ribbon areas from $x=a$ to $x=b$.

$2.$ The integration will not be immediate. You can let $\sqrt{2}\,x=\tan\theta$ or $\sqrt{2}\,x=\sinh t$.

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