Surface area using double integrals

[Carnivroar]

Find the surface area of the triangle with vertices (0,0) (L, L) (L,-L)

I know I have to take the double integrals of f(x,y) but I have no idea what f(x,y) is supposed to be!

 

[Delphi51]

Why do you need an integral to find the area of a triangle?
Why not good old A = bh/2 ?

 

[thepatient]

A surface area in the form of a double integral will be in the form:

∫∫dA

 

[thepatient]

But Delphi's answer is just as good too if you simply would like to know a double integral and not use any calculus. :]

 

[thepatient]

Oops, I mean area, not double integral.

 

[Carnivroar]

A surface area in the form of a double integral will be in the form:

∫∫dA

what's dA? I wasn't there during the lecture so I'm lost. I looked through my textbooks, online, and I can't find anything. All I think I know is that I need a f(x, y).

And no I can't use bh/2. But it's good because I know my answer from the double integral has to be something like bh/2 so I can verify my answer.

 

[thepatient]

dA would be dxdy. You will first have to integrate with respect to x or y, then integrate with respect to the second variable. If your double integrals limits of integration are just numbers, will give you the area of a rectangle. Your limits of integration can also be functions as well though.

 

[Carnivroar]

dA would be dxdy. You will first have to integrate with respect to x or y, then integrate with respect to the second variable. If your double integrals limits of integration are just numbers, will give you the area of a rectangle. Your limits of integration can also be functions as well though.

but what do i integrate?

∫∫(?)dydx

 

[thepatient]

dA is basically a very small element of the area and depends on the type of coordinate system you use as well. If you just take a very small rectangle to be dA, then your dA will be dx dy. If you're using polar coordinates and your dA is a strip defined by the radius and the angle, then dA = r dr d(theta).

When you integrate, you hold every other variable that you're not integrating over as a constant. For example, let's say you have the double integral of y+x dydx.

∫x = 0..2 ∫y = 0..3 y+x dy dx

Integrating with respect to y, holding all other variables constant:

∫x=0..2 y^2/2 + xy | y = 0..3 dx =

∫x = 0..2 4.5 + 3x dx =

4.5x + 3x^2/2 x = 0..2 =

9 + 6 = 15.

 

[Carnivroar]

dA is basically a very small element of the area and depends on the type of coordinate system you use as well. If you just take a very small rectangle to be dA, then your dA will be dx dy. If you're using polar coordinates and your dA is a strip defined by the radius and the angle, then dA = r dr d(theta).

When you integrate, you hold every other variable that you're not integrating over as a constant. For example, let's say you have the double integral of y+x dydx.

∫x = 0..2 ∫y = 0..3 y+x dy dx

Integrating with respect to y, holding all other variables constant:

∫x=0..2 y^2/2 + xy | y = 0..3 dx =

∫x = 0..2 4.5 + 3x dx =

4.5x + 3x^2/2 x = 0..2 =

9 + 6 = 15.

I know how to do double integrals but I don't know what should come here ∫∫(??)dydx

The image is a triangle

It looks like this

 

[thepatient]

In your case, you will have:

∫^x=b_x=a ∫^{y = g(x)}_{y = h(x)}dydx

Or you can have it in the form of:
∫^y=d_y=c ∫^{x= g(y)}_{x = h(y)}dxdy

But in your case I think it would be easier to do the first. The integrand will just be 1 dydx.

So you will have to find an upper limit function and a lower limit function. Once you have those functions and you integrate, then you integrate with respect to your other variable and use your limits a to b.

You may want to graph the triangle. Your upper limit will be the equation of a line of form y = mx + b and your lower limit will be when y = 0. I'm not sure if I'm allowed to just give out the answer. XD Give it a try and look up some example problems on the book too. :]

 

[thepatient]

Ooh in that case, you will have your limits as y = mx +b and y = -wx + h, where m and w are separate slopes of each line and b and h are different y intercepts. You will still be integrating simply over 1 dydx.

 

[thepatient]

In summary, a double integral of 1 gives you an area, a triple integral of 1 gives you a volume.

 

[Carnivroar]

positive slope = 1x + 0 = x
negative slope = -1x + 0 = -x

not sure if I'm sticking them in the right place

 x   x
 ∫   ∫ dydx
-x   0

I got x²

I guess that's the same thing as L²

I plugged in some numbers on the bh/2 formula and they seem to agree

wow so simple.

am I correct now?

 

[Carnivroar]

Actually I think it should be the other way around

 x   x 
∫   ∫ dydx
0   -x

Still comes out to x₂

 

[thepatient]

Very good. :] Now imagine if you were to cut the triangle right in half through the x axis, and fit the bottom half on the top left of the top half. You have a rectangle with area of L^2.

 

[Carnivroar]

Very good. :] Now imagine if you were to cut the triangle right in half through the x axis, and fit the bottom half on the top left of the top half. You have a rectangle with area of L^2.

Yep, thanks!

 

[thepatient]

Mathematically, you would probably want to change your x limits from 0 to x to limits of 0 to x prime, or another letter.