Due to the symmetry of the paraboloid about the z-axis, you can treat this is a surface of revolution. Consider the curve $y=x^2$ , with $1\le x\le2$ , and revolve it about the y-axis. The area of the resulting surface is then

$\displaystyle2\pi\int_1^2x\sqrt{1+(y')^2}\,\mathrm dx=2\pi\int_1^2x\sqrt{1+4x^2}\,\mathrm dx=\frac{(17^{3/2}-5^{3/2})\pi}6$

But perhaps you'd like the surface integral treatment. Parameterize the surface by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k$

with $1\le u\le2$ and $0\le v\le2\pi$ , where the third component follows from

$z=x^2+y^2=(u\cos v)^2+(u\sin v)^2=u^2$

Take the normal vector to the surface to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial u}=-2u^2\cos v\,\vec\imath-2u^2\sin v\,\vec\jmath+u\,\vec k$

The precise order of the partial derivatives doesn't matter, because we're ultimately interested in the magnitude of the cross product:

$\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|=u\sqrt{1+4u^2}$

Then the area of the surface is

$\displaystyle\int_0^{2\pi}\int_1^2\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\int_0^{2\pi}\int_1^2u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$

which reduces to the integral used in the surface-of-revolution setup.

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3 years ago