Question

Surface Integrals for fun. \[\int\int_S xyz dS\] \(S\) is the cone with parametric equations \(x=u\cos(v)~,~ y=u\sin(v)~,~z=u~,~0\le u \le 1~,~ 0\le v \le \pi/2\)

Answer 1

since they've given us \(x~,y~,z\), we could write our parametric equation for our surface \(\vec r(u,v)\) I'm guessing.

Answer 2

No you were right, why delete it?

Answer 3

one sec deleted by mistake

Answer 4

\[\vec r(u,v) =\langle u\cos(v)~,u\sin(v)~,~u\rangle\]

Answer 5

\[\iint_S f(x,y,z)~ \mathrm{d}S = \iint_D f(\vec{r}(u,v))~\| \vec{r}_u \times \vec{r}_v\|~\mathrm{d}A\]

Answer 6

To check your answer, I got the integral evaluated to
\[
\frac{1}{5 \sqrt{2}}
\]
unless I did some small computation error.

Answer 7

Yeah, it was a small computational error.

Answer 8

im getting the same..

Answer 9

\[\frac{1}{10}\sqrt{2}\]

Answer 10

\[\left|\vec r_u \times \vec r_v\right| = \left[\begin{matrix}\hat i & \hat j & \hat k \\ \cos(v) & \sin(v) & 1\\ -\sin(v) & \cos(v) & 0 \end{matrix}\right] \\ =\langle 0-\cos(v)~,~ -(0+\sin(v))~,~ \cos^2(v) +\sin^2(v)\rangle = \langle -\cos(v)~,~ -\sin(v)~,~ 1\rangle \]

Answer 11

Oh, forgot to take the magnitude.

Answer 12

\[\left|\vec r_u \times \vec r_v\right| = \left|\begin{matrix}\hat i & \hat j & \hat k \\ \cos(v) & \sin(v) & 1\\ -\color{red}{u}\sin(v) & \color{red}{u}\cos(v) & 0 \end{matrix}\right| \\ \]

Answer 13

Oh, because w.r.t v.

Answer 14

yes i got the same !
just a notation thing : use vertical bars for determinant. square brackets refer to a matrix, not determinant

Answer 15

matrix A :
\[A = \begin{bmatrix}a&b\\c&d \end{bmatrix}\]

determinant of matrix A :
\[|A| = \begin{vmatrix}a&b\\c&d \end{vmatrix}\]

Answer 16

\[\left|\vec r_u \times \vec r_v\right| = \left|\begin{matrix}\hat i & \hat j & \hat k \\ \cos(v) & \sin(v) & 1\\ -u\sin(v) & u\cos(v) & 0 \end{matrix}\right| \\=\langle 0-u\cos(v)~,~ -|(0+u\sin(v))~,~ u\cos^2(v) +u\sin^2(v)\rangle \\ = \langle -u\cos(v)~,~ -u\sin(v)~,~ u\rangle \\ =\left | \vec r_u \times \vec r_v \right| = \sqrt{u^2\cos^2(v) +u^2\sin^2(v)+u^2} =u\sqrt{2}\]

Answer 17

looks great!

Answer 18

\[\iint_S f(x,y,z)~ \mathrm{d}S = \iint_D f(\vec{r}(u,v))~\| \vec{r}_u \times \vec{r}_v\|~\mathrm{d}A\] \[x=u\cos(v)~,~y=u\sin(v)~,~z=u\]\[\iint_D (u\cos(v))(u\sin(v))(u)\cdot u\sqrt{2}dudv = \sqrt{2}\iint_D (u^3\cos(v)\sin(v))dudv \]

Answer 19

you should get u^4 in integrand..

Answer 20

\[\iint_D (\color{red}{u}\cos(v))(\color{red}{u}\sin(v))(\color{red}{u})\cdot \color{red}{u}\sqrt{2}dudv = \sqrt{2}\iint_D (u^{\color{red}{4}}\cos(v)\sin(v))dudv\]

Answer 21

Oh I missed a u :(

Answer 22

miscounted -.-

Answer 23

bounds and the differential are not agreeing.. check once

Answer 24

\[\sqrt{2} \int_0^1\int_0^{\pi/2} \left(u^4 \cos(v)\sin(v)\right)dvdu\]

Answer 25

wolfram it :)

Answer 26

:P

Answer 27

I can split the integrals, then on the v integral just use a u-sub, right?

Answer 28

\[\sqrt{2}\left[\int_0^1 u^4du \int_0^{\pi/2} \cos(v)\sin(v)dv\right]\]

Answer 29

let \(u=\sin(v)~,~ du = \cos(v)dv\)\[\sqrt{2} \left[\left[\frac{1}{5}u^5\right]_0^1 \cdot \left[\int_0^1 udu\right]\right] = \sqrt{2}\left[\frac{1}{5}\cdot \left. \left(\frac{1}{2}u^2\right|_0^1 \right) \right] =\sqrt{2} \cdot\frac{1}{5} \cdot \frac{1}{2} =\frac{1}{10}\sqrt{2} \]

Answer 30

There we go ) Just a fun, refreshing surface integral before I headed to sleep :P

Answer 31

Wow! you have evaluated manually and managed to arrive at the correct answer! good job :)

Answer 32

Haha, thanks!! It's good practice.