\(\)
Completely different integrals
Triple integrals
$$ \iiint_R f\,dV \nonumber $$
Integrates a scalar quantity over a 3D region. The shape of the region is described in the integral’s bounds.
Volume element
 In Cartesian coordinates: \(dV = dx\,dy\,dz\)
 In cylindrical coordinates: \(dV = dz\,r\,dr\,d\theta\)
 In spherical coordinates: \(dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta\). Remember
 \(\rho\) is the distance from the origin
 \(\phi\) is the angle with the positive \(z\)axis, where \(0\lt\phi\lt\pi\)
 \(\theta\) is the angle around the \(z\)axis, where \(0\lt\phi\lt 2\pi\)
Set the bounds of integration

For Cartesian and cylindrical coordinates:
 For \(\int dz\), for a fixed point in the \(xy\)plane, find the bounds for \(z\). The bottom and top surface of your solid as a function of \((x,y)\).
 For the outer integrals, look at the solid from above, to see its projection on the \(xy\)plane. Set up a double integral in rectangular or polar coordinates for the bounds of \(x\) and \(y\). If you did \(z\) first, the inner bounds are given by bottom and top, and the outer ones are given by looking at the shadow of the region.

For spherical coordinates:
 For \(\int d\rho\), it is like you shoot a ray from the origin to space. You want to know what part of you beam is in the solid. You want to solve for the value of \(\rho\) when you enter and when you leave the solid.
 For \(\int d\phi\), sketch the \(rz\)plane to find the bounds for \(\rho\).
 For \(\int d\theta\), ask yourself what values of \(\theta\) will I be inside my region.
Applications
 Mass, \( \rm m = \iiint_R\delta\,dV \)
 Average value of \(f\), \( \bar f=\frac{1}{\rm{m}}\iiint_R f\,\delta\,dV \)
 Moment of inertia around \(x\)axis, \( I_x = \iiint_R (y^2+z^2)\,\delta\,dV \)
 Gravitational attraction on mass at origin, \( F_z = \frac{GmM}{a^2} \)
Surface integrals
$$ \iint_S \vec F\cdot\hat n\,dS \nonumber $$
Integrates a vector quantity over a 2D surface. The shape of the region is described in the integral’s bounds.
Area element
 In Carthesian coordinates: \(dA = dy\,dx\):
 In polar coordinates: \(dA = r\,dr\,d\theta\):
Formulas for \(\hat n\,dS\) in various settings

Special surfaces
 Horizontal plane: \( \hat n = \left\langle 0,0,\pm1\right\rangle \) and \( dS = dx\,dy \)
 Vertical plane \(\newcommand{\parallelsum}{\mathbin{\!/\mkern5mu/\!}} \parallelsum\) \(yz\)plane: \( \hat n = \left\langle \pm 1,0,0\right\rangle \) and \( dS = dy\,dz \)
 Sphere centered at origin: \( \hat n = \pm\frac{\left\langle x,y,z\right\rangle}{a} \) and \( dS = a^2 \sin\phi\,d\phi\,d\theta \), where \(a\) is the radius.
 Cylinder centered at \(z\)axis: \( \hat n = \pm\frac{\left\langle x,y,0\right\rangle}{a} \) and \( dS = a\,dz\,d\theta \), where \(a\) is the radius.

General case
 Given \(z = z(x,y)\): \( \hat n\,dS = \pm\left\langle z_x, z_y, 1\right\rangle\,dx\,dy \)
 Given the normal vector \(\hat N\): \( \hat n\,dS = \pm\frac{\vec N}{\vec N\cdot\hat k}\,dx\,dy \)
That will turn it into a regular double integral.
You then set up the bounds in terms of the integration variables.
Line integrals
$$ \int_C \vec F\cdot\,d\vec r \nonumber $$
Integrates a vector quantity over a 1D curve. The shape of the curve is described in the integral’s bounds.
Evaluate as $$ \int_C \vec F\cdot\,d\vec r = \int_C P\,dx + Q\,dy + R\,dz \nonumber $$
Approach
 Parameterize the curve \(C\) to express \(x,y,z\) in terms of a single (parametric) variable.
 Solve the single integral.
Bridges between the integrals
Each of these theorems relates a quantity with a certain number of integral signs to a quantity with one more integral sign.
Triple integral – Surface integral
The divergence theorem connects these.
For a closed surface \(S\), you can replace the flux integral with a triple integral over the region inside. $$ \iint_S\vec F\cdot\hat n\,dS = \iiint_D (\rm{div}\,\vec F)\,dV \nonumber $$
Here \(\vec F\) is a vector field, and \(\rm{div}\,\vec F\) is a function that relates to the vector field.
Surface integral – Line integral
Stokes’ theorem connects these.
I can replace a line integral on closed curve \(C\), with a double integral on the surface \(S\) $$ \oint_C \vec F\cdot d\vec r = \iint_S (\nabla\times\vec F)\cdot\hat n\,dS \nonumber $$
It relates a line integral for field \(\vec F\) to a surface integral from another field, \(\nabla\times\vec F\).
You compute it as any other surface integral. You find the formula for \(\hat n\cdot dS\), do the dotproduct, substitute and evaluate. The calculation doesn’t know it came from a curl, and is the same as any other flux integral.
Line integral – Function
The line integral for a gradient of a function is equal to the change in value of the function. Given \(\vec F\) with curl \(=0\), find potential
The fundamental theorem of calculus says $$ f(P_1) – f(P_0) = \int_C (\nabla f)\cdot d\vec r \nonumber $$
The line integral for the vector field given by the gradient of the function, is equal to the change in value of the function.