Question

Calculus 3: Part of the surface \(y=4x+z^2\) that lies between planes \(x=0~, ~x=1~,~z=0~,~z=1\)

Answer 1

Lets see it

Answer 2

I have parametricized my equations as : \( x=x ~ , ~ y=4x+z^2 ~ , ~ z=z\)

Answer 3

Yes that is correct

Answer 4

That isn't what I am looking for @Pico33

Answer 5

No. I am finding the surface Area. \[A = \int\int_D \sqrt{1+(f_x)^2 +(f_y)^2}dA\]

Answer 6

Ahh ok

Answer 7

Well, you've pretty much set it up.

Answer 8

I'm not really sure where to go with this problem now..

Answer 9

Is that really the curl? The tangent vector is \(\langle f_x, -1, f_z\rangle\) then?

Answer 10

This example problem is telling me I need to use the cross product to figure out my vector, yes, is that what you are referring to?

Answer 11

Yeah, I'm talking about the curl of the tangent vectors of the surface

Answer 12

I'm confused.

Answer 13

I don't quite get where you are stuck?

Answer 14

\[
f(x,z) = 4x+z^2
\]

Answer 15

Oh now I see where \(\langle f_ x, -1, f_z\rangle\) comes into the picture.

Answer 16

It's because i'm given a function in terms of the component \(y \hat j\)?

Answer 17

Yeah, when we have a parametrization where two of the variables are in our x,y,z coordinate system, it end up having that normal vector.

Answer 18

It's sort of special benefit you get from using the same coordinate system.

Answer 19

I see. Makes sense. I'll try working it out.

Answer 20

Do I need to start by taking the cross product?

Answer 21

Have you done surface integrals before? Should I start from the beginning?

Answer 22

Yeah.. I was just working through the homework trying to understand as I went along.

Answer 23

The same way that a path integral over the function \(f(x,y,z) = 1\) gives arc length, a surface integral over this function will give us surface area. Surface integrals are basically a special notation.

Answer 24

\[
\iint_Sf ~dS
\]Is a special notation. \(S\) stands for some surface we integrate over. \(f\) is the scalar function we are integrating over. Lastly \(dS\) is the change in surface area. Since we want surface area we set \(f=1\).

Answer 25

First thing we will do is parametrize \(S\). We need two free variables to parametrize a surface. In general we will find a parametrization like \(\Phi(u,v) = \langle x(u,v), y(u,v), z(u,v)\rangle\). Where we bind the parametrization variables with: \(u_1 < u < u_2, v_1<v<v_2\).

Answer 26

And finally \(\large dS = \left|\frac{\partial \Phi}{\partial u}\times \frac{\partial \Phi}{\partial v}\right|du~dv\)

Answer 27

Oh I see. It's becoming a little bit more clear now.

Answer 28

\[
\iint_Sf ~dS = \int_{u_1}^{u_2}\int_{v_1}^{v_2}f~ \left|\frac{\partial \Phi}{\partial u}\times \frac{\partial \Phi}{\partial v}\right|du~dv
\]

Answer 29

Now, \(f\) is a function of \(x,y,z\), and we compose it with \(\Phi\) to get a function of \(u,v\)

Answer 30

Alright.

Answer 31

In our case, we get surface area so \(f=1\).

Also when we parametrize such that only one variable \(x,y,z\) is a function of the other two, we already know what the normal vector will be

Answer 32

We know it would be \(\langle f_x,f_y,-1 \rangle\)

Answer 33

That is if we had \(z=f(x,y)\). The variable that becomes a function of the other gets turned to \(-1\), the others are replaced with their partial derivatives of the parametrization function

Answer 34

Ahhh, yess.

Answer 35

\[
|\langle f_x,f_y,-1\rangle| = \sqrt{\langle f_x,f_y,-1\rangle\cdot \langle f_x,f_y,-1\rangle} = \sqrt{f_x^2+f_y^2+1}
\]

Answer 36

Well, you can easy see this if you actually try to find the curl yourself

Answer 37

\[
\begin{vmatrix}
\mathbf i&\mathbf j&\mathbf k\\
\frac{\partial}{\partial x}x &\frac{\partial}{\partial x}y&\frac{\partial}{\partial x}f(x,y)\\
\frac{\partial}{\partial y}x &\frac{\partial}{\partial y}y&\frac{\partial}{\partial y}f(x,y)\\
\end{vmatrix}
=
\begin{vmatrix}
\mathbf i&\mathbf j&\mathbf k\\
1 &0&f_x\\
0 &1&f_y\\
\end{vmatrix} = \langle f_x,f_y,-1\rangle
\]

Answer 38

My usage of \(f\) here is not the function we are integrating over, it's the parametrization: \[
\Phi (x,y) = \langle x,y,f(x,y)\rangle
\]

Answer 39

So going back to the actual problem at hand...

Answer 40

\[

\Phi(x,y)=\langle x ,4x+z^2 , z\rangle

\]

Answer 41

Oh, I was rereading your posts on how the variable that becomes the function turns into -1, and how you got z = -1. I understand that now.

Answer 42

And \[
dS = \sqrt{\Bigg(\frac{\partial }{\partial x}\bigg(4x+z^2\bigg)\Bigg)^2+1+\Bigg(\frac{\partial }{\partial z}\bigg(4x+z^2\bigg)\Bigg)^2}dx~dz
\]

Answer 43

All together:\[
S=\iint_SdS = \int_0^1\int_0^1\sqrt{\Bigg(\frac{\partial }{\partial x}\bigg(4x+z^2\bigg)\Bigg)^2+1+\Bigg(\frac{\partial }{\partial z}\bigg(4x+z^2\bigg)\Bigg)^2}dx~dz
\]Where \(S\) is the surface area of our parametrized surface

Answer 44

And you got the limits of integration because \(0 \le x \le 1\) and \(0\le z \le 1\)

Answer 45

yes

Answer 46

Basically, when you parametrize with the same coordinate system, you don't have to compute the curl of the tangent vectors because it will follow the pattern

Answer 47

Yeah, I see that now.

And so if this equation was in terms of x, instead of y or z. We would have \(\langle -1, f_y, f_z\rangle\)

Answer 48

But it's a special case. You shouldn't forget how you'd do a surface integral with a custom parametrization

Answer 49

\[
\Phi(u,v) = \langle \cos u\sin v,\sin u\sin v,\cos v\rangle
\]

Answer 50

\[
\begin{vmatrix}
\mathbf i&\mathbf j&\mathbf k\\
\frac{\partial}{\partial u}\cos u \sin v &\frac{\partial}{\partial u}\sin u \sin v&\frac{\partial}{\partial u}\cos v\\
\frac{\partial}{\partial v}\cos u \sin v &\frac{\partial}{\partial v}\sin u \sin v&\frac{\partial}{\partial v}\cos v\\
\end{vmatrix}

\]

Answer 51

< sin(u)'sin(v) * (cos(v))' - (0)*sin(u)sin(v)' , ........>

Answer 52

\[
\int_C ds
\]

Answer 53

\[
\int_C dr
\]

Answer 54

\[
\iint_S dS
\]

Answer 55

\[
\int_Cf~dr
\]

Answer 56

\[
\iint _Sf~dS
\]

Answer 57

\[
\int_C\mathbf F\cdot d\mathbf r
\]

Answer 58

\[
\iint_S\mathbf F\cdot d\mathbf S
\]

Answer 59

\[\int_0^1 \sqrt{17+4z^2}dz\]

Answer 60

\[\int_0^1 \sqrt{4(\frac{17}{4} +z^2)}dz\]

Answer 61

\[2\int_0^1 \sqrt{\frac{17}{4} +z^2}dz\]

Answer 62

\[
1^2+\tan^2\theta = \sec^2\theta
\]

Answer 63

\[
\sqrt{17/4}^2+(\sqrt{17/4}\tan\theta)^2=(\sqrt{17/4}\sec\theta)^2
\]

Answer 64

\[
\sin^2\theta+\cos^2=1
\]

Answer 65

\[
\cos^2\theta=1-\sin^2\theta
\]

Answer 66

\[
\tan^2\theta+1=\sec^2\theta
\]

Answer 67

\[
1-x^2\\
1+x^2\\
x^2-1
\]

Answer 68

\[
1-x^2 \to 1-\sin^2\theta=\cos^2\theta\\
1+x^2\to 1+\tan^2\theta=\sec^2\theta \\
x^2-1 \to \sec^2\theta-1=\tan^2\theta
\]

Answer 69

\[
z = \sqrt{17/4}\tan\theta
\]

Answer 70

\[
a^2-x^2 \to a^2-a^2\sin^2\theta=a^2\cos^2\theta\\
a^2+x^2\to a^2+a^2\tan^2\theta=a^2\sec^2\theta \\
x^2-a^2 \to a^2\sec^2\theta-a^2=a^2\tan^2\theta
\]