41.1 Area of a parametric surface

41.1 Area of a parametric surface

1. A partition of $S$ into patches will correspond to a partition of $D$ (in the $u$-$v$ plane) into small rectangles.

The dimensions of the rectangles in $D$ will be $\Delta u\Delta v$.

41.1 Area of a parametric surface

2. Let one of the edges of a single patch be defined from parameter values $(u, v)$ to $(u+ \Delta u, v)$.

41.1 Area of a parametric surface

Using Pythagoras' law in three dimensions, we can approximate the length of this edge as $\text{length}$ $\approx \sqrt{\left(\Delta x\right)^2 + \left(\Delta y\right)^2 + \left(\Delta z\right)^2}$ $\displaystyle = \left(\sqrt{\left(\frac{\Delta x}{\Delta u}\right)^2 + \left(\frac{\Delta y}{\Delta u}\right)^2 + \left(\frac{\Delta z}{\Delta u}\right)^2} \right) \Delta u$ $\approx \norm{\r_u}\Delta u,$ where in this case we have used $$\Delta x = x(u + \Delta u, v) -x(u, v)$$ etc (i.e., the change is only in $u$). Similarly, for an edge of patch running from parameter values $(u, v)$ to $(u, v + \Delta v)$ the length of that edge will be approximately $||\r_v||\Delta v.$

41.1 Area of a parametric surface

At the corner of the patch corresponding to parameter values $(u, v),$ we can define the two vectors $\r_u\Delta u$ and $\r_v\Delta v$ which form two sides of a parallelogram, the side lengths of which coincide with our approximations to the lengths of the edges of the patch.

41.1 Area of a parametric surface

The vector $\left(\r_u~\Delta u\right) \times \left(\r_v~\Delta v\right)$ is normal to the surface (and hence the tangent plane) at that point. Its magnitude gives the area of the parallelogram we use to approximate the area of the patch $\Delta S$. We then have $$\Delta S \approx \norm{\r_u \times \r_v} ~\Delta u ~\Delta v.$$ Adding these approximations for each patch in $S$ gives us an approximation to the area of $S$: area of $S$ $\displaystyle\approx \sum _i \Delta S_i$ $\displaystyle= \sum_i \norm{\r_{u_i} \times \r_{v_i}} ~\Delta u_i ~\Delta v_i.$

41.1 Area of a parametric surface

4. Finally taking the limit as $\Delta u, \Delta v \to 0$ we obtain $$\text{surface area} = \iint_S dS = \iint_D \norm{\r_u \times \r_v} ~du ~dv.$$

41.1.1 Application: find the surface area of the paraboloid $z=1-x^2-y^2$ for $z\geq 0$

Let's begin with the parametrisation of $S$. Consider $x$ and $y$ as parameters, so $$\r(x,y) = x~\i + y~\j + \left(1-x^2-y^2\right)~\k$$ $x$ and $y$ are bounded by intersection curve of $$z=1-x^2-y^2\quad \text{and} \quad z=0.$$ That is, $x^2+y^2=1$. So the domain for $x$ and $y$ is given by $x^2+y^2\leq 1$.

41.1.1 Application: find the surface area of the paraboloid $z=1-x^2-y^2$ for $z\geq 0$

Thus $$\r_x = \i-2x~\k, \quad \r_y = \j-2y~\k,$$ and $\r_x\times\r_y = \left| \begin{array}{ccc} \i & \j & \k\\ 1 & 0 & -2x \\ 0 & 1 & -2 y \end{array} \right|$ $= 2x~\i + 2y ~\j + \k$. Since $||\r_x\times\r_y||=\sqrt{4x^2+4y^2+1},$ then $$dS = ||\r_x\times\r_y|| ~dx~dy = \sqrt{4x^2+4y^2+1}~dx~dy.$$

41.1.1 Application: find the surface area of the paraboloid $z=1-x^2-y^2$ for $z\geq 0$

$\Rightarrow \text{Area}$ $\displaystyle = \iint_S dS$ $\displaystyle = \iint\limits_{x^2+y^2\leq 1} ||\r_x\times\r_y|| ~dx~dy$ $\displaystyle = \iint\limits_{x^2+y^2\leq 1} \sqrt{4x^2+4y^2+1}~dx~dy$ $\displaystyle = \int_0^{2\pi} \int_0^1 \sqrt{4r^2+1}~r~dr~d\theta$ $\displaystyle = \int_0^{2\pi} \left[ \bigg. \frac{1}{8}~\frac{2}{3} \left(4r^2+1\right)^{3/2} \bigg|_0^1 \right] d\theta$ $\displaystyle = \int_0^{2\pi} \left[ \frac{5 \sqrt{5} -1}{12} \right] d\theta$ $\displaystyle =\frac{\pi}{6} \left(5 \sqrt{5} - 1\right).$

41.2 More on calculating surface integrals, applications

41.2 More on calculating surface integrals, applications

41.2 More on calculating surface integrals, applications

41.2 More on calculating surface integrals, applications

41.2.1 The function $T (x, y, z) = x^2 + y^2 + z^2 + 4$

gives the temperature at any point $(x, y, z)$ on the surface of a solid hemisphere of radius 1 centred at the origin, defined for $z\geq 0$. Find the average temperature over the surface

Here we have that $\displaystyle \text{Average temp.} = \frac{\displaystyle\iint_S T~dS}{\text{area of }S}\qquad \qquad \qquad \qquad$ $\qquad \;\;= \frac{\displaystyle\iint_{S_1} T~dS + \iint_{S_2} T~dS }{\displaystyle\iint_{S_1} ~dS + \iint_{S_2} ~dS },$ where $S_1$ is the base and $S_2$ is the top.

41.2.1 The function $T (x, y, z) = x^2 + y^2 + z^2 + 4$

gives the temperature at any point $(x, y, z)$ on the surface of a solid hemisphere of radius 1 centred at the origin, defined for $z\geq 0$. Find the average temperature over the surface

For $S_1$: Here we have a disk in the $xy$-plane of radius $1$. Then the parametrisation is $$\r(x,y) = x~\i+y~\j+0~\k\quad (x^2+y^2\leq 1).$$ Then $\r_x = \i$, $\r_y = \j$ and $$dS = \norm{\r_x\times \r_y }~dx~dy= dx~dy.$$ So $\displaystyle \iint_{S_1} T~dS = \iint_{S_1} \left( x^2+y^2+z^2+4\right)dx~dy$ or $\displaystyle \iint_{S_1} T~dS = \iint_{S_1} \left( x^2+y^2+(0)^2+4\right)dx~dy$ That is $\displaystyle \iint_{S_1} T~dS = \iint_\limits{x^2+y^2\leq 1} \left( x^2+y^2+4\right)dx~dy$.

41.2.1 The function $T (x, y, z) = x^2 + y^2 + z^2 + 4$

gives the temperature at any point $(x, y, z)$ on the surface of a solid hemisphere of radius 1 centred at the origin, defined for $z\geq 0$. Find the average temperature over the surface

For $S_1$: Using polar coordinates $$x = r\cos\theta , \quad y = r \sin \theta$$ with $0\leq r \leq 1$ and $0\leq \theta \leq 2 \pi$, we have $\displaystyle \iint_{S_1} T~dS$ $\displaystyle = \iint_\limits{x^2+y^2\leq 1} \left( x^2+y^2+4\right)dx~dy$ $\displaystyle = \int_0^1 \int_0^{2\pi} \left(r^2 \cos ^2 \theta + r^2 \sin^2 \theta + 4\right)r ~dr~d\theta$ $\displaystyle = \int_0^1 \int_0^{2\pi} \left(r^2 + 4\right)r ~dr~d\theta$ $\displaystyle = \frac{9 \pi}{2}$

41.2.1 The function $T (x, y, z) = x^2 + y^2 + z^2 + 4$

gives the temperature at any point $(x, y, z)$ on the surface of a solid hemisphere of radius 1 centred at the origin, defined for $z\geq 0$. Find the average temperature over the surface

For $S_1$: Using polar coordinates $$x = r\cos\theta , \quad y = r \sin \theta$$ with $0\leq r \leq 1$ and $0\leq \theta \leq 2 \pi$, we have $\displaystyle \iint_{S_1} T~dS$ $\displaystyle = \frac{9 \pi}{2}$. Finally $\;\;\displaystyle \iint_{S_1}~dS$ $\displaystyle = \text{Area of the disk of radius 1}$ $=\pi (1)^2$ $= \pi$.

41.2.1 The function $T (x, y, z) = x^2 + y^2 + z^2 + 4$

gives the temperature at any point $(x, y, z)$ on the surface of a solid hemisphere of radius 1 centred at the origin, defined for $z\geq 0$. Find the average temperature over the surface

For $S_2$: Now we have a half sphere of radius 1. Then the parametrisation is $$\r(\theta,\phi ) = \cos\theta \sin \phi~\i+ \sin\theta \sin \phi ~\j+\cos \phi~\k$$ where $0\leq \theta \leq 2 \pi$ and $0 \leq \phi \leq \pi/2$. Then $\quad \r_{\theta} = -\sin\theta \sin \phi~\i+ \cos\theta \cos \phi ~\j+0~\k$ $\quad \r_{\phi} = \cos\theta \cos \phi~\i+ \sin\theta \cos \phi ~\j- \sin \phi~\k$

41.2.1 The function $T (x, y, z) = x^2 + y^2 + z^2 + 4$

gives the temperature at any point $(x, y, z)$ on the surface of a solid hemisphere of radius 1 centred at the origin, defined for $z\geq 0$. Find the average temperature over the surface

41.2.1 The function $T (x, y, z) = x^2 + y^2 + z^2 + 4$

gives the temperature at any point $(x, y, z)$ on the surface of a solid hemisphere of radius 1 centred at the origin, defined for $z\geq 0$. Find the average temperature over the surface

$\norm{\r_{\theta}\times\r_{\phi} }$ $=\sqrt{ \cos^2 \theta \sin ^4 \phi + \sin^2 \theta \sin ^4 \phi + \sin^2 \phi \cos^2 \phi }$ $=\sqrt{ \left( \cos^2 \theta + \sin^2 \theta \right) \sin ^4 \phi + \sin^2 \phi \left(1- \sin ^2\phi\right) }$ $=\sqrt{ \sin^4 \phi + \sin^2 \phi - \sin^4\phi }$ $=\sqrt{ \sin^2 \phi }$ $= \sin \phi$ So $$\iint_{S_2} T~dS = \iint_{S_2}\left(x^2+y^2+z^2+4\right)dS$$

41.2.1 The function $T (x, y, z) = x^2 + y^2 + z^2 + 4$

gives the temperature at any point $(x, y, z)$ on the surface of a solid hemisphere of radius 1 centred at the origin, defined for $z\geq 0$. Find the average temperature over the surface

$\displaystyle \iint_{S_2} T~dS$ $\displaystyle = \iint_{S_2}\left(x^2+y^2+z^2+4\right)dS$ $\displaystyle = \iint_{S_2}\left(1+4\right)dS$ $\displaystyle = 5 \iint_{S_2}dS$ $\displaystyle = 5 \left(\text{Surface area of the half sphere}\right)$ $=\displaystyle 5\left(\frac{1}{2}\right) \left(4 \pi \right)\left(1 \right)^2$ $= 10 \pi.$ And $\displaystyle \iint_{S_2} ~dS$ $\displaystyle = \text{Surface area of the half sphere}$ $=$ $\displaystyle\, 2 \pi$.

41.2.1 The function $T (x, y, z) = x^2 + y^2 + z^2 + 4$

gives the temperature at any point $(x, y, z)$ on the surface of a solid hemisphere of radius 1 centred at the origin, defined for $z\geq 0$. Find the average temperature over the surface

Therefore $\displaystyle \text{Average temp.} = \frac{\displaystyle\iint_{S_1} T~dS + \iint_{S_2} T~dS }{\displaystyle\iint_{S_1} ~dS + \iint_{S_2} ~dS }$ $\;\;\;\quad \displaystyle = \frac{\displaystyle \frac{9\pi}{2}+ 10 \pi}{\displaystyle \pi + 2\pi}$ $\displaystyle = \frac{29}{6}\;$.