20241105

You end up with a pyramid, with a square base on the $xz$-plane. Note that $z = 1-x$ is a plane parallel to the $y$-axis, and $y=1-x$ is a plane parallel to the $x$-axis. We want the region in the first octant (all variables are nonnegative) that is bounded by these two planes.

Here's a solution with Mathematica.

R = ParametricRegion[{x, y, z}, {{x, 0, 3}, {y, 0, 1 - x}, {z, 0, 1 - y}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]

The $x$-coordinate of the centroid is $$\bar x = \frac{\iiint_D x dV}{\iiint_D dV} = \frac{\ds\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-y}xdzdydx}{\ds\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-y}dzdydx}.$$

The volume integral is 1/3, and I'll jump straight to Mathematica to compute it.

• The first option focus on the iterated nature of the integral.
• The second option is streamlined.
• The third option puts the bounds as variables at the top of your code. This allows for using the bounds in many ways at the same time. The Evaluate[] command may be needed at times to force Mathematica to use the variables you have defined.

Integrate[Integrate[Integrate[1, {z, 0, 1 - y}], {y, 0, 1 - x}], {x, 0, 1}]
Integrate[1, {x, 0, 1}, {y, 0, 1 - x}, {z, 0, 1 - y}]

outerB = {x, 0, 1}
middleB = {y, 0, 1 - x}
innerB = {z, 0, 1 - y}
volume = Integrate[1, outerB, middleB, innerB]

R = ParametricRegion[{x, y, z}, Evaluate[{outerB, middleB, innerB}]];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
centroid = Integrate[{x, y, z}, outerB, middleB, innerB]/volume

The region described by $-3\leq x\leq 3$ and $0\leq y\leq \sqrt{9-x^2}$ is the upper half of a disc of radius 3. Bounds in polar coordinates are $0\leq \theta\leq \pi$ and $0\leq r\leq 3$. Note that $x^2+y^2=r^2$, which means $e^{-x^2-y^2} = e^{-r^2}$. Also remember the Jacobian is $|r|$. We now have $$\int_{-3}^{3}\int_{0}^{\sqrt{9-x^2}}e^{-x^2-y^2}dydx = \int_{0}^{\pi}\int_{0}^{3}e^{-r^2}rdrd\theta.$$

Note that $\int_{0}^{3}e^{-r^2}rdr$ does not contain $\theta$, which means this will become a constant once we've finished the inside integral. This gives $$\int_{0}^{\pi}\int_{0}^{3}e^{-r^2}rdrd\theta = \left(\int_{0}^{3}e^{-r^2}rdr\right)\int_{0}^{\pi}d\theta.$$ The point is that we just separated the problem into two different integrals.

• The integral with respect to $\theta$ yields $\pi$.
• The integral with respect to $r$ requires a substitution (such as $u = -r^2$) and yields $\frac{1}{2}(1-e^{-9})$.

Overall, this gives $\int_{-3}^{3}\int_{0}^{\sqrt{9-x^2}}e^{-x^2-y^2}dydx = \frac{\pi}{2}(1-e^{-9}).$

With Mathematica, the polar integral will evaluate exactly, whereas the rectangular one will not. To get exact answers, sometimes you MUST change to a different coordinate system. Both quickly evaluate when using a numerical approximation.

Integrate[Exp[-r^2] r, {theta, 0, Pi}, {r, 0, 3}]
Integrate[Exp[-x^2 - y^2], {x, -3, 3}, {y, 0, Sqrt[9 - x^2]}]

NIntegrate[Exp[-r^2] r, {theta, 0, Pi}, {r, 0, 3}]
NIntegrate[Exp[-x^2 - y^2], {x, -3, 3}, {y, 0, Sqrt[9 - x^2]}]