# Why do we need cross product in surface integral?

Why do we need cross product in surface integral? I can't quite understand the idea of doing integration over surface. For example, if we have a surface $f$, the steps to take surface integral is to reparametrize the surface and then find the cross product and then take the integral $$\int_s \int_t f(x(s,t)) \left\|\frac{\partial x}{\partial s} \times \frac{\partial x}{\partial t}\right\| \,ds\,dt.$$ The reasoning is that we want to "sum up" small region of tangent place weighted by each "chunk" of the surface $f$ (the cross product).

I cannot understand why in this case I have to have the extra cross product to be able to sum up all the "chunks" while in the normal double integral case, I do not need to have it and still is able to sum up all the "chunks". What am I misunderstanding here? What is the difference between these 2 cases that necessitates the cross product?

• One thing to note is that the area of a parallelogram is the magnitude of the cross product of the vectors that form it's sides... So what's the area element of a surface? Naturally, it's the magnitude of the cross product of the partial derivatives, hence the term in your integral. As noted in the answers given, this is $1$ for integration in the $x-y$ plane. I can give you a derivation of these notions if you wish. – superckl Nov 8 '17 at 17:15
• Your notation doesn't make sense: $$\int_s \int_t f(x(s,t)) \left\|\frac{\partial x}{\partial s} \times \frac{\partial x}{\partial t}\right\| \,ds\,dt \text{ ?}$$ This is $$\int_s \left( \int_t f(x(s,t)) \left\|\frac{\partial x}{\partial s} \times \frac{\partial x}{\partial t}\right\| \,ds \right) \,dt$$ Thus the inside integral is $$\int_t f(x(s,t)) \left\|\frac{\partial x}{\partial s} \times \frac{\partial x}{\partial t}\right\| \,ds$$ i.e. $$\int_t \cdots\cdots\, ds$$ and the outside one is $$\int_s \cdots \cdots \, dt.$$ What does that mean? $\qquad$ – Michael Hardy Nov 13 '17 at 15:43
• It is a typo. Will correct it later. – user10024395 Nov 14 '17 at 1:16

In the normal double integral case where the surface is the $x-y$ plane, you are effectively using a parametrization $x(t,u) = (t,u,0)$ only you sensibly call the parameters $(x,y)$ rather than $(t,u).$ Then your cross product is $$(1,0,0) \times (0,1,0) =(0,0,1)$$ which has length one so the factor is just one, which explains why you never saw it.
The factor is there to translate the the area $dsdt$ of a small piece of parameter space into the area of the corresponding small piece of the surface. In the simple case above where the surface and the parameter space are basically identified, the area of the corresponding piece of the surface is just $dxdy$ and no conversion is needed.
The area of a parallelogram spanned by $a$ and $b$ is $a\times b$. In the surface area integral case, we have $a=\partial x/\partial u$ and $b=\partial x/\partial v$. In the standard area integral in the Cartesian plane, we have $a=\hat{i}$ and $b=\hat{j}$ so the area is $\lVert\hat{i}\times\hat{j}\rVert=\lVert\hat{k}\rVert=1.$