Which coordinate system fits this integral?
Rectangular, cylindrical, or spherical — read the boundary before you set up bounds.
Look at what block of variables the region's boundary equation combines. If x and y only ever appear as x^2+y^2, that block is crying out to become r^2 (cylindrical). If x,y,z only ever appear together as x^2+y^2+z^2, spherical collapses it to one variable, \rho^2. If neither happens, stay in rectangular — converting adds work a plain box integral doesn't need.
The three systems
A point is located by (r,\theta,z), where (r,\theta) are its usual polar coordinates in the xy-plane and z is the same rectangular height as before. x = r\cos\theta, \qquad y = r\sin\theta, \qquad z = z r^2 = x^2+y^2, \qquad \tan\theta = \frac{y}{x} The volume element picks up the same factor polar coordinates already contribute in the plane. dV = r\,dz\,dr\,d\theta
Example
A point is located by (\rho,\theta,\phi). Here \rho\ge 0 is the straight-line distance from the origin, \theta is the same polar angle as in cylindrical coordinates, and \phi\in[0,\pi] is the angle down from the positive z-axis. x = \rho\sin\phi\cos\theta, \qquad y = \rho\sin\phi\sin\theta, \qquad z = \rho\cos\phi \rho^2 = x^2+y^2+z^2 The volume element is dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta
Example
Show proof
Build the box [\rho,\rho+d\rho]\times[\theta,\theta+d\theta]\times[\phi,\phi+d\phi] one edge at a time. Moving \rho by d\rho with \theta,\phi fixed sweeps a straight radial edge of length d\rho. Moving \phi by d\phi with \rho,\theta fixed sweeps an arc of a great circle of radius \rho, length \rho\,d\phi. Moving \theta by d\theta with \rho,\phi fixed sweeps an arc around the z-axis at horizontal radius \rho\sin\phi (the same r a cylindrical slice at that height would have), length \rho\sin\phi\,d\theta. The three edges are mutually perpendicular, so the box's volume is their product: d\rho\cdot(\rho\,d\phi)\cdot(\rho\sin\phi\,d\theta) = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta.
The decision itself
Section
Choosing a Coordinate System
Formula
The Jacobian
For a transformation x=x(u,v),\ y=y(u,v), the Jacobian is the determinant of the matrix of first partial derivatives. \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[4pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix} = \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u} Its absolute value \left|\dfrac{\partial(x,y)}{\partial(u,v)}\right| is the factor by which the transformation stretches or shrinks area near (u,v). A small rectangle of area du\,dv in the uv-plane maps to a region of area approximately \left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|du\,dv in the xy-plane. The same construction extends to three variables as a 3\times3 determinant of first partials, \dfrac{\partial(x,y,z)}{\partial(u,v,w)}.
Whatever system you pick, this is the scale factor that makes the substitution rigorous, not optional.
Show proof
Near (u_0,v_0), the transformation's best linear approximation sends the edge vector (du,0) to \left(\frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}\right)du and the edge vector (0,dv) to \left(\frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}\right)dv — the two columns of the Jacobian matrix, scaled. The image of the small du\times dv rectangle is (to first order) the parallelogram spanned by those two vectors, and the area of a parallelogram spanned by (a,b) and (c,d) is |ad-bc|, exactly the absolute value of that 2\times2 determinant.
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