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The question given is the following:

Let $\Omega$ be the domain bounded by the paraboloid z = 4 - ($x^2$ + $y^2$) and the plane $z = 0$.

Let $f$ be the scalar field $f(x,y,z) = ax + by + cz$, where a, b and c are constants. Find $∫_{\Omega}$ $f$ $dV$.

I know that the answer to this is 32$\pi$$c$/3, I calculated it myself.

The lecturer made an aside comment for an exercise; that being

"Without doing the integral, how does the answer depend on a, b and c"

I'm very confused about this, and have no idea, any help?

Sam Connell
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    Look for jacobian in polar coordinates. For example, you can find it here https://en.wikipedia.org/wiki/Polar_coordinate_system#Integral_calculus_(area) – Mario G Jan 05 '22 at 13:11
  • Integration is sometimes a method of calculating an area, but not usually. If you wanted the area of that region you should set $f(r,\theta)=1$, but you have a different function here. – David K Jan 05 '22 at 14:08
  • The result is a linear function of a,b and c by additivity of the integral. By symmetry the contributions linear in a and b vanish. – Thomas Jan 15 '22 at 22:03
  • @Thomas why do they vanish? I understand the volume is symmetric about the z-axis, but why would they vanish? – Sam Connell Jan 15 '22 at 22:34
  • Take the x component and make a symmetry sending x->x'=-x and everything else unchanged . The integrand corresponding to the x component is odd w.r.t. this transfrmation but the domain is invariant. So the two contributions cancel. You can formalize with a change of variables to see that after this transformation the integral is equal to minus itself. – Thomas Jan 16 '22 at 02:57

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The volume is symmetric about the z axis so the $ax$ and $by$ terms in the field, which will be positive on one side and negative on the other cancel themselves out. So they can be nulled. So the problem is the same as $f(x,y,z)=cz$.

Because the field only depends on z, the field is constant for a particular value of z. Also, cross sections of the paraboloid at a particular z value are circles. We could integrate the area of the circles from z=0 to z=4 multiplied by the field to get the answer. But we are avoiding integration so let's just scale the area by the field and see what happens.

First let's get an expression for the cross section area from the original paraboloid equation.

$z = 4 - (x^{2} + y^{2})$

$4-z = (x^{2} + y^{2})$

$\pi(4-z) = \pi(x^{2} + y^{2}) = \pi r^{2}$ where $r^{2} = x^{2} + y^{2}$

Okay so now that we have the expression for the area of the cross sections, lets multiply the area by the field.

$\pi(4-z) * f(x,y,z) = \pi(x^{2} + y^{2})$

Substitute in the new expression for the field.

$\pi(4-z) * cz = \pi(x^{2} + y^{2})$

So now it looks like we can have an $x^{2}$, $y^{2}$, and $z^{2}$ terms so this equation is looking like that of a sphere or an ellipsoid. Lets do some manipulation.

$(4-z) * cz = (x^{2} + y^{2})$

$c(4z-z^{2}) = (x^{2} + y^{2})$

$-c(z^{2}-4z+4-4) = (x^{2} + y^{2})$

$-c(z^{2}-4z+4)+4c = (x^{2} + y^{2})$

$-c(z-2)^{2}+4c = (x^{2} + y^{2})$

$4c = x^{2} + y^{2}+c(z-2)^{2}$

$1 = \frac{x^{2}}{(2\sqrt{c})^{2}} + \frac{y^{2}}{(2\sqrt{c})^{2}}+\frac{(z-2)^{2}}{(2)^{2}}$

This is now in the form of an ellipsoid. The standard form for an ellipsoid is:

$1 = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}$

The formula for the volume of an ellipsoid is

$V = \tfrac{4}{3}\pi abc$

So the volume of our ellipsoid is

$V = \tfrac{4}{3}\pi (2\sqrt{c})*(2\sqrt{c})*(2) = 32\pi c/3$

So this is the final result

$∫_{\Omega}$ $f$ $dV = 32\pi c/3$

I admit I would not have found this route of solving without integration if your question did not say it was possible. I think this problem is interesting and it makes me wonder if a field can be thought of as a fourth dimension.

C. A.
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  • Okay thank you, he also said to try it for the function z = 4 - root(x^2 + y^2); I know that gives 64cpi/3, but is it similar logic. In addition, I'm a bit confused on why they cancel each other out. Obviously this is the case with many normal integrals in two dimensional spaces, e.g. if you integrate x from -1 to 1 the areas are equal so they cancel, but why would this apply in 3D for volume integrals? Bit confused on that front – Sam Connell Jan 15 '22 at 22:33
  • Interesting that it works for that function with the root. I may try it. Your analogy to two dimensions for why the $ax$ and $by$ terms cancel out is correct. In 3D you are summing (integrating) over the volume but you still have the same number of points in space that have ax terms that are less than zero on the -x side of the x-axis as there are points in space with positive ax terms on the +x side of the x-axis. – C. A. Jan 16 '22 at 15:35
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This is mostly an elaboration of Thomas's comment.

By linearity, the integral is equal to $$ I=a\int_\Omega x\;dxdydz+b\int_\Omega y\;dxdydz+c\int_\Omega z\;dxdydz\;. $$

Now consider the integral $\displaystyle\int_\Omega -x\;dxdydz$ and the change of variable $$ u=-x, y=y, z=z\;. $$ Denote this linear transformation as $\varphi$. Then $|\det(D\varphi)|=1$ and $\varphi(\Omega)=\Omega$. So by the change of variable formula, you have $$ \int_\Omega -x\;dxdydz=\int_{\varphi(\Omega)} u\;dudydz=\int_\Omega x\;dxdydz $$ which implies that $\displaystyle\int_\Omega x\;dxdydz=0$. Similarly, you have $\displaystyle\int_\Omega y\;dxdydz=0$.

Consequently, $$ I=c\int_\Omega z\;dxdydz $$ which is a multiple of the constant $c$.