# Do I have to include the Jacobian when using this form of the surface integral?

I know that when we are given a surface $P$ by its parametrization with respect to $t$ and $s$$\iint_{P} \left\lVert \frac{\partial P}{\partial t} \times \frac{\partial P}{\partial s} \right\rVert dP$$ Right now, no additional Jacobian is necessary, because the vector product will cater for it. I have two more questions that keep troubling me: 1. If I decided to change the parametrization before calculating the cross product, for example, to switch from the Cartesian to the Cylindrical coordinate system, would it still be enough to rely on the vector product, or additional Jacobian would be necessary? 2. If I decided to use this formula instead: $$\iint_P \sqrt{1+\left(\frac{\partial P}{\partial s}\right)^2+\left(\frac{\partial P}{\partial t} \right)^2}dP$$ What about the Jacobian right now? Words are cheap and so perhaps let's take a look at this simple example: Find the surface area of the cone$x^2 + y^2 =z^2\$

Parametrization: $$(x, y, \sqrt{x^2+y^2})$$
In cylindrical: $$(r, \theta, r)$$

Would I have to use the Jacobian with this integral?

• Yes, if you change parametrization you will need to uniformize by the inverse of the Jacobian for the surface area and flux/curl/divergence integrals as per usual. – John Samples Jun 17 '18 at 9:23