I have the following math question:
Find the surface area of the part of the paraboloid $ z=5-(x^2 + y^2)$ that lies between
the planes $z=0$ and $z=1$.
So far i have computed
$\sqrt{fx^2+fy^2+1}$
to be
$\sqrt{4x^2+4y^2+1}$
and have come to:
$\int \int \sqrt{4x^2+4y^2+1} dx dy$
over a region, R. Is this the right way to go about setting up this equation? I know the region over which i must integrate is the reflection of $ z=5-(x^2 + y^2)$ ( with $z=0$ and $z=1$) on the xy plane, which is hard for me to find. I still have issues with xyz sketching. Can someone please help me sketch this/ give me steps on sketching?
Best Answer
\begin{align} {\rm d}A &= \left\vert {\partial\vec{r} \over \partial x}\times{\partial\vec{r} \over \partial y} \right\vert \,{\rm d}x\,{\rm d}y = \left\vert \left(\hat{x} - 2x\,\hat{z}\right)\times\left(\hat{y} - 2y\,\hat{z}\right) \right\vert \,{\rm d}x\,{\rm d}y = \left\vert\,\hat{z} + 2y\,\hat{y} + 2x\,\hat{x}\,\right\vert\,{\rm d}x\,{\rm d}y \\[3mm]&= \sqrt{4\left(x^{2} + y^{2}\right) + 1\, }\ \,{\rm d}x\,{\rm d}y \end{align}
\begin{align} & \\[5mm] A &= \int_{0\ <\ z\ <1}\sqrt{4\left(x^{2} + y^{2}\right) + 1\,}\ \ {\rm d}x\,{\rm d}y = \int_{2}^{\sqrt{5}}\sqrt{1 + 4\rho^{2}\,}\ \ \rho\ {\rm d}\rho \int_{0}^{2\pi}{\rm d}\theta \\[3mm]&= \pi\int_{4}^{5}\sqrt{1 + 4z\,}\ {\rm d}z = \pi\,\left\lbrack% {\left(1 + 4\times 5\right)^{3/2} \over 6} - {\left(1 + 4\times 4\right)^{3/2} \over 6} \right\rbrack = {\large{1 \over 6}\,\pi\,\left(21^{3/2} - 17^{3/2}\right)} \end{align}