Surface Area Multiple Integrals problem

  • Thread starter Zonda
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  • #1
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Hi,
I need some help on these problems. I'm not sure what to do.

1 Find the area of the plane with vector equation r(u, v) =< 1+v, u-2v, 3-5u+v> that is given by 0<u<1, 0<v<1.

So far, I took the partial derivatives with respect to u and v. I don't know if I was supposed to or not and I'm really stuck there.

Since I'm trying to find surface area a relevant equation may be S = (double integral) SQRT(1 + (partial derivative of u)^2 + (partial derivative of v)^2) dA

2. The part of the surface z = 1 + 3x +2y^2 that lies above the triangle with vertices (0,0), (0,1), (2,1)

Relevant equation should be same as above.

I think I got the answer but I'm not sure and would really appreciate some help if I got it wrong.
Here's how I did it:
partial x: 3 Partial y: 4y
The boundary of the triangle is x=2y
so it took the integral from 0 to 1 and from 0 to 2y SQRT(10 + 16y^2)dx dy
and I got the integral from 0 to 1 times 2y SQRT(10 +16y^2) dy
then I took that integral and got (2/3)*(2/32)*SQRT(10 + 16Y^2)^(3/2) from 0 to 1
My final answer is (1/24)(26)^(3/2) - (1/24)(10)^(3/2)

Is this correct?
 

Answers and Replies

  • #2
arildno
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Remember that when you parametrize a surface, tangent vectors are gained by differentiating with respect to the two independent variables.
Let us take the general case.

Suppose you have a surface S parametrized as
[tex]\vec{S}(u,v)=(X(u,v),Y(u,v),Z(u,v))[/tex]
Then, two infinitesemal tangent vectors to this surface are given by:
[tex]d\vec{t}_{u}=\frac{\partial\vec{S}}{\partial{u}}du,d\vec{t}_{v}=\frac{\partial\vec{S}}{\partial{v}}dv[/tex]
The differential area spanned by these two vectors lies on S, and its area is given by:
[tex]dA=||d\vec{n}_{S}||=||d\vec{t}_{u}\times{d}\vec{t}_{v}||=||\frac{\partial\vec{S}}{\partial{u}}\times\frac{\partial\vec{S}}{\partial{v}}||dudv[/tex]
And the total area of S is then gained by summing together all dA's (i.e, by integration).

Thus, yes, you have thought correctly on 1.
 
  • #3
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The answer I got for number 1 is SQRT (107), is this correct?

What about number 2?
 
  • #4
arildno
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Hmm..I have to retract what I said about you having thought correctly on 1.

What you have written is totally meaningless!

Follow what I wrote in my post:
[tex]\frac{\partial\vec{r}}{\partial{u}}=\vec{j}-5\vec{k},\frac{\partial\vec{r}}{\partial{v}}=\vec{i}-2\vec{j}+\vec{k}[/tex]
Thus, the norm of the cross product is:
[tex]||\frac{\partial\vec{r}}{\partial{u}}\times\frac{\partial\vec{r}}{\partial{v}}||=\sqrt{62}[/tex]
and since both u and v varies between 0 and 1, [itex]\sqrt{62}[/itex] is the correct answer.
 
  • #5
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I still get SQRT (107). Are you sure about SQRT(62)?
 
  • #6
arildno
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Hmm..perhaps not. Made a too fast calculation. Sorry about that. square root of 107 it is.
 
  • #7
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Thanks! What about number 2?
 
  • #8
HallsofIvy
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Hmm..I have to retract what I said about you having thought correctly on 1.

What you have written is totally meaningless!

Follow what I wrote in my post:
[tex]\frac{\partial\vec{r}}{\partial{u}}=\vec{j}-5\vec{k},\frac{\partial\vec{r}}{\partial{v}}=\vec{i}-2\vec{j}+\vec{k}[/tex]
Thus, the norm of the cross product is:
[tex]||\frac{\partial\vec{r}}{\partial{u}}\times\frac{\partial\vec{r}}{\partial{v}}||=\sqrt{62}[/tex]
and since both u and v varies between 0 and 1, [itex]\sqrt{62}[/itex] is the correct answer.
???
[tex]\left|\begin{array}{ccc}i & j & k \\ 0 & 1 & -5 \\ 1 & -2 & 1\end{array}\right|= i\left|\begin{array}{cc}1 & -5 \\ -2 & 1\end{array}\right|- j\left|\begin{array}{cc}0 & -5 \\1 & 1\end{array}\right|+ k\left|\begin{array}{cc} 0 & 1 \\ 1 & -2\end{array}\right|[/tex]
= -9 i- 5j - k
which has length [itex]\sqrt{81+25+1}= \sqrt{107}[/itex]
 
  • #9
arildno
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Indeed, it is.
However, as I made the mental calculation (-5)*(-2)=-5, so that the i'th component became -6 rather than -9, I got 62 as my radicand rather than 107.
 
Last edited:
  • #10
HallsofIvy
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You are starting to take after me. That is NOT a good sign!
 
  • #11
HallsofIvy
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Yes, the answer to (1) is [itex]\sqrt{107}[/itex]

For (2): The part of the surface z = 1 + 3x +2y^2 that lies above the triangle with vertices (0,0), (0,1), (2,1)

Pretty much the same thing. You can write this as x= u, y= v, z= 1+ 3u+ 2v2, basically using x and y as parameters, so that r= ui+ vj+ (1+ 3u+ 2v2)k. Find ru, rv[/sup] and take their cross product. Integrate over the triangle (which has sides x= 0, y= 1, and y= (1/2)x).
 
  • #12
AKG
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Another way for 1:

The surface you're looking for is a parallelogram with vertices at r(0,0), r(1,0), r(0,1) and r(1,1). So two of its non-parallel sides are given by r(1,0)-r(0,0) and r(0,1)-r(0,0). Find the norm of the cross product of these two vectors:

|| [r(1,0)-r(0,0)] x [r(0,1)-r(0,0)] ||

and this will be the desired area.
 
  • #13
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Thanks for your help!
 

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