Thread: Find area of a portion of a plane (defined with vector function)

1. Find area of a portion of a plane (defined with vector function)

Find the area of the portion of $x + 2y + 4z = 10$ in the first octant.
So, first, we need to find vector function for the surface.

A few point that lie on the plane are $\langle 2,1,1 \rangle$, $\langle 0,5,0 \rangle$, $\langle 2,4,0 \rangle$.

Now we can find two vectors parallel to the plane by subtracting a few points:
$\langle 0,5,0 \rangle - \langle 2,4,0 \rangle = \langle -2,1,0 \rangle$
$\langle 0,5,0 \rangle - \langle 2,1,1 \rangle = \langle -2,4,-1 \rangle$

So our vector function is: $\vec r(u,v)= \langle 2,1,1\rangle +u\langle -2,1,0\rangle + v\langle -2,4,-1\rangle=\langle 2-2u-2v,1+u+4v,1-v \rangle$.

To find area we'll also need partial derivatives of $\vec r$:

$\vec r_u = \langle -2,1,0\rangle$ and $\vec r_v = \langle -2,4,-1 \rangle$

So area of the surface is
\begin{align*}
\int_{a}^{b} \int_{c}^{d} ||\vec r_u \times \vec r_v||\,du\,dv&=\int_{a}^{b} \int_{c}^{d} ||\langle -2,1,0\rangle \times \langle -2,4,-1 \rangle||\,du\,dv\\
&=\int_{a}^{b} \int_{c}^{d} ||\langle -1,-2,-6 \rangle||\,du\,dv\\
&=\int_{a}^{b} \int_{c}^{d} \sqrt{(-1)^2 + (-2)^2 + (-6)^2}\,du\,dv\\
&=\int_{a}^{b} \int_{c}^{d} 41\,du\,dv

\end{align*}

Well, it seems to be the procedure in the book, but I'm not sure this was correct? Also I'm not sure how to find the boundaries for the integrals? Can someone help?

2. Re: Find area of a portion of a plane (defined with vector function)

Your method makes it awfully difficult. I hesitate to help you on this one though because my method is so different.

As far as finding the limits. You should certainly know that the lower limit for both $x$ and $y$ is $0$.

To find the upper limit of $x$ we note that it occurs when $y$ and $z$ are both $0$, i.e. at $x=10$.

Then for $y$, given an $x$, we know the max $y$ occurs at $\dfrac {10-x}{2}$

That's all there is to it.

I'm attaching a mathematica sheet that shows how I would do this. You integrate the differential area of the surface across those limits.

if $A$ is defined by $f(x,y)$ then $dA = \sqrt{1+f_x^2+f_y^2}$

The bit at the bottom is a cross check using the formula for area of a triangle given 3 points. As you can see they agree.

3. Re: Find area of a portion of a plane (defined with vector function)

Hum, thank you. So the problem is to find $u$ and $v$ so that they make $x$ go from $0$ to $1$, and from $0$ to $(10-x)/2$?

4. Re: Find area of a portion of a plane (defined with vector function)

The point $\displaystyle (2,1,1)$ doesn't lie on the plane.

Right .

6. Re: Find area of a portion of a plane (defined with vector function)

Ah, okay. I found a method. There's a simpler vector function for the plane, and with it it's easier to write the limits, using romsek's hint.

Basically, $\vec r(u,v)=\langle u,v, \frac{10-u-2v}{4}\rangle$.

And then $\int_{0}^{10} \int_{0}^{\frac{10-v}{2}}||\vec r_u \times \vec r_v||\,dv\,du$ gives the correct answer.

7. Re: Find area of a portion of a plane (defined with vector function)

Originally Posted by maxpancho
Ah, okay. I found a method. There's a simpler vector function for the plane, and with it it's easier to write the limits, using romsek's hint.

Basically, $\vec r(u,v)=\langle u,v, \frac{10-u-2v}{4}\rangle$.

And then $\int_{0}^{10} \int_{0}^{\frac{10-v}{2}}||\vec r_u \times \vec r_v||\,dv\,du$ gives the correct answer.
This is perfect.

As an exercise.

let

$\vec{r}=<u,v,f(u,v)>$

apply your method to obtain a formula for the differential area and see that it is the formula I used.

So your method is just one step back from where I started and is probably easier to extend to higher dimensions.

8. Re: Find area of a portion of a plane (defined with vector function)

Originally Posted by maxpancho
Ah, okay. I found a method. There's a simpler vector function for the plane, and with it it's easier to write the limits, using romsek's hint.

Basically, $\vec r(u,v)=\langle u,v, \frac{10-u-2v}{4}\rangle$.

And then $\int_{0}^{10} \int_{0}^{\frac{10-v}{2}}||\vec r_u \times \vec r_v||\,dv\,du$ gives the correct answer.
You know you could also pick $\displaystyle \vec r(u,v)=\langle 10 - 2u - 4v,u,v \rangle$. It would avoid the fractions.

9. Re: Find area of a portion of a plane (defined with vector function)

Originally Posted by maxpancho
Ah, okay. I found a method. There's a simpler vector function for the plane, and with it it's easier to write the limits, using romsek's hint.

Basically, $\vec r(u,v)=\langle u,v, \frac{10-u-2v}{4}\rangle$.

And then $\int_{0}^{10} \int_{0}^{\frac{10-v}{2}}||\vec r_u \times \vec r_v||\,dv\,du$ gives the correct answer.
Originally Posted by romsek
This is perfect.

As an exercise.

let

$\vec{r}=<u,v,f(u,v)>$

apply your method to obtain a formula for the differential area and see that it is the formula I used.

So your method is just one step back from where I started and is probably easier to extend to higher dimensions.
actually it's not quite perfect.

The inner upper limit should be $\dfrac {10-u}{2}$ not $\dfrac{10-v}{2}$ as you have it written.

Below is a shameless plug for Mathematica since it's just about the only piece of software that I've ever truly enjoyed using.

Your problem in 2 lines.

10. Re: Find area of a portion of a plane (defined with vector function)

Ah, yeah, confusing u's and v's .

11. Re: Find area of a portion of a plane (defined with vector function)

I have a few of these to solve. Here's another one:

Find the area of the portion of $2x + 4y + z = 0$ inside $x^2 + y^2 = 1$.
Any hints for finding limits?

PS Actually, I got an idea just now. Will try to solve it.

12. Re: Find area of a portion of a plane (defined with vector function)

I think the simplest way to answer this question is to used "Hero's formula": the area of a triangle with side lengths a, b, and c, is $\displaystyle \sqrt{s(s- a)(s- b)(s- c)}$ where s is half the perimeter, (a+ b+ c)/2. The vertices of the triangle lie on the three axes. When y= z= 0, x= 10 so one vertex is (10, 0, 0). When x= z= 0, 2y= 10 so another vertex is (0, 5, 0). When x= y= 0, 4z= 10 so (0, 0, 5/2) is the third vertex.
The lengths of the three sides are $\displaystyle \sqrt{125}= 5\sqrt{5}$, $\displaystyle \sqrt{25+ 25/4}= \sqrt{125}/2= (5/2)\sqrt{5}$, and $\displaystyle \sqrt{100+ 25/4}= (5/2)\sqrt{17}$. $\displaystyle s= (15/4)\sqrt{5}+ (5/4)\sqrt{17}$.

13. Re: Find area of a portion of a plane (defined with vector function)

What's the area of the triangle whose vertices are: (10, 0, 0), (0, 5, 0), and (0, 0, 5/2) ?

One approach: It's a triangle with sides of length $\displaystyle 5 \sqrt{5}, \frac{5}{2} \sqrt{17}, \frac{5}{2} \sqrt{5}$, so use Heron's Formula: Heron's Formula -- from Wolfram MathWorld

Another approach: it's 1/2 the area of the parallelogram formed by the vectors $\displaystyle \vec{v_1} = -10 \hat{i} + 5 \hat{j}$ and $\displaystyle \vec{v_2} = -10 \hat{i} + \frac{5}{2} \hat{k}$, so it's $\displaystyle \frac{1}{2} \big| \ \vec{v_1} \times \vec{v_2} \ \big|$.

14. Re: Find area of a portion of a plane (defined with vector function)

what did you get?

ps. while it's true the 3 point area formula is easiest it's fairly clear he's working on methodology for computing surface integrals.

15. Re: Find area of a portion of a plane (defined with vector function)

I get $\displaystyle \vec{v_1} \times \vec{v_2} = \frac{25}{2} \hat{i} + 25 \hat{j} + 50 \hat{k}$

so that

$\displaystyle \big| \ \vec{v_1} \times \vec{v_2} \ \big| = \big| \ \frac{25}{2} \hat{i} + 25 \hat{j} + 50 \hat{k} \ \big| = \frac{25}{2} \big| \ \hat{i} + 2 \hat{j} + 4 \hat{k} \ \big| = \frac{25}{2} \sqrt{1 + 4 + 16}$$\displaystyle = \frac{25 \sqrt{21}}{2}. Thus the area of the triangle is \displaystyle \frac{25 \sqrt{21}}{4}. I get the same result when applying Heron's Formula: \displaystyle s = \frac{15\sqrt{5} + 5 \sqrt{17}}{4}, s - a = \frac{-5\sqrt{5} + 5 \sqrt{17}}{4}, s - b = \frac{15\sqrt{5} - 5 \sqrt{17}}{4}$$\displaystyle , s - c = \frac{5\sqrt{5} + 5 \sqrt{17}}{4}$

Thus Area $\displaystyle = \sqrt{\left( \frac{15\sqrt{5} + 5 \sqrt{17}}{4} \right) \cdot \left( \frac{-5\sqrt{5} + 5 \sqrt{17}}{4} \right) \cdot \left( \frac{15\sqrt{5} - 5 \sqrt{17}}{4} \right) \cdot \left( \frac{5\sqrt{5} + 5 \sqrt{17}}{4} \right)}$

$\displaystyle = \left( \frac{5}{4} \right)^2 \sqrt{\left( 3\sqrt{5} + \sqrt{17} \right) \cdot \left( -\sqrt{5} + \sqrt{17} \right) \cdot \left( 3\sqrt{5} - \sqrt{17} \right) \cdot \left( \sqrt{5} + \sqrt{17} \right)}$

$\displaystyle = \frac{25}{16}\sqrt{\left( (3\sqrt{5})^2 - (\sqrt{17})^2 \right) \cdot \left( (\sqrt{17})^2 - (\sqrt{5})^2 \right)} = \frac{25}{16}\sqrt{ (45 - 17) (17 - 5)}$

$\displaystyle = \frac{25}{16}\sqrt{ (28) (12)} = \frac{25}{16}\sqrt{ (4 \cdot 7) (4 \cdot 3)} = \frac{25}{16}\sqrt{ 4^2 \cdot 21 }$

$\displaystyle = \frac{25}{4}\sqrt{ 21 }$.

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