Double Integrals: Area or Volume?

[mkkrnfoo85]

I'm reading in my Calculus book, and I see (I may see wrongly) that a double integral can describe both an Area and a Volume. Is that true? If that's true, how do I know when the Double Integral is describing an Area or a Volume? Thanks.

 

[Moo Of Doom]

Well, it's sort of similar to how a single integral can be used to find both area and volume (and length, as well).

 

[dextercioby]

The typical triple integral describing a volume of a certain domain $\mathcal{D}\subseteq \mathbb{R}^{3}$ is

$$V_{D}=\iiint_{\mathcal{D}} \ dV$$

Daniel.

 

[Galileo]

It depends on your problem. An integral doesn't necessarily describe something geometric, but it can be used to calculate surface areas, for example:

$$\iint_D \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial g}{\partial y}\right)^2}dA$$
or
$$\iint_D 1 dA$$

or volumes:

$$\iint_R f(x,y)dA$$

It depends on the problem.

 

[mkkrnfoo85]

ok thanks.

 

[Theelectricchild]

$$\iint_D \sqrt{1+\left(\frac{\partial g}{\partial x}\right)^2+\left(\frac{\partial g}{\partial y}\right)^2}\partial A$$

correct?

 

[dextercioby]

Yeah,it was a typo by Galileo,but we usually write $z=z\left(x,y\right)$ when we indicate the equation of a surface in $\mathbb{R}^{3}$ explicitely .

Daniel.

 

[Theelectricchild]

Yeah you are correct, that was one of my problems when I was first learning surface integrals... whenever I would get stuck, I would jump to the conclusion that I could simply use that general equation to solve the problem--- but it only works in specific cases. (Or when you're scrambling on a final!)