How to Find the Area of a Surface in the First Octant?

[wrongusername]

Homework Statement

Find the area of the surface: The part of the plane 3x + 2y + z = 10 that lies in the first octant

Homework Equations

$$A=\iint_{D}|\vec{r}_{u}\times\vec{r}_{v}|dA$$

The Attempt at a Solution

I parametrized the plane:

$$\vec{r}\left(u,v\right)=<u,v,6-3u-2v>$$

and found the partial derivatives of u and v
$$\vec{r}_{u}=<1,0,-3>$$

$$\vec{r}_{v}=<0,1,-2>$$

then got this for their cross product:
$$\vec{r}_{u}\times\vec{r}_{v}=<3,2,1>$$

$$|\vec{r}_{u}\times\vec{r}_{v}|=\sqrt{14}$$

I think u goes from 0 to 2 because that's the highest and lowest values for u before the plane exits the first octant. This is also why I think the integral of v goes from 0 to 3. I obtain this answer:
$$\int_{0}^{2}\int_{0}^{3}\sqrt{14}dvdu=6\sqrt{14}$$

when the actual answer is half of that. What have I done wrong?

 

[tiny-tim]

Hi wrongusername!

(I haven't ploughed through all your equations, but …)

shouldn't the u limits depend on v (or vice versa)?

 

[LCKurtz]

$$\vec{r}\left(u,v\right)=<u,v,6-3u-2v>$$

And shouldn't that 6 be a 10?

 

[wrongusername]

Hi wrongusername!

(I haven't ploughed through all your equations, but …)

shouldn't the u limits depend on v (or vice versa)?

Hi tiny-tim

I thought a surface in 3D was supposed to be parameterized with 2 variables u and v? (whereas a line integral in 2D was parameterized with just one variable t)

And shouldn't that 6 be a 10?

Sorry, I copied the question down wrong

It should be 3x + 2y + z = 6. Don't know where I got that 10 from.

 

[LCKurtz]

Hi tiny-tim

I thought a surface in 3D was supposed to be parameterized with 2 variables u and v? (whereas a line integral in 2D was parameterized with just one variable t)

That's right. But your u and v are just new names for x and y. If you draw the first octant portion of the plane by locating its three intercepts, you will see the (x,y) domain is a triangle. Your limits on your integral describe a 2 by 3 rectangle.

 

[wrongusername]

That's right. But your u and v are just new names for x and y. If you draw the first octant portion of the plane by locating its three intercepts, you will see the (x,y) domain is a triangle. Your limits on your integral describe a 2 by 3 rectangle.

Oh! I see. Since the domain is a triangle, its area is half of the rectangle I had integrated.

Thank you!

 

[tiny-tim]

Oh! I see. Since the domain is a triangle, its area is half of the rectangle I had integrated.

Hold it!

Yes, it is a triangle, and therefore half the area of the rectangle, but that's not what you're supposed to be learning from this question.

You need to understand how one of the limits of a double integral can depend on one of the variables, and you clearly don't understand that yet.

Go back to your professor (or your book) and find how to integrate over irregular areas.

 

[wrongusername]

Hold it!

Yes, it is a triangle, and therefore half the area of the rectangle, but that's not what you're supposed to be learning from this question.

You need to understand how one of the limits of a double integral can depend on one of the variables, and you clearly don't understand that yet.

Go back to your professor (or your book) and find how to integrate over irregular areas.

I think I do understand

The domain is the triangle with vertices (2, 0), (0, 3), and (0, 0) in the xy plane; thus, I integrate like this

$$\intop_{0}^{2}\intop_{0}^{3-1.5x}\sqrt{14}dydx=\intop_{0}^{2}\sqrt{14}y|_{0}^{3-1.5x}dx=\sqrt{14}\intop_{0}^{2}3-1.5xdx=\sqrt{14}\left[3x-0.75x^{2}\right]_{0}^{2}$$

$$=\sqrt{14}\left(6-3\right)=3\sqrt{14}$$

and get half the answer I first got. It's just that seeing it as a triangle with half the area of the rectangle was a lazy way to understand what I did wrong

Thank you for your help!

 

[LCKurtz]

And hopefully you realize that had there been a variable density involved, taking half the answer over the rectangle would most likely have given the wrong answer. Dangerous shortcut.

 

[wrongusername]

And hopefully you realize that had there been a variable density involved, taking half the answer over the rectangle would most likely have given the wrong answer. Dangerous shortcut.

I know, but in this case there was just a simple constant to integrate, so I thought it was safe