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You should obtain $y=(cd-af)z/(ae-bd)$. A similar computation obtains $x = (bf-ce)z/(ae-bd)$. We can pick $z=ae-bd$, which then gives $$\vec u\times \vec v = (x,y,z)=(bf-ce,cd-af,ae-bd).$$

There are two options, based on how you decide to set up the bounds. These options are $$m = \iint_R dm = \iint_R \delta dA = \int_{0}^{ \sqrt{8} }\int_{x^2}^{8}xy^2dydx = \int_{0}^{8} \int_{0}^{\sqrt{y}}xy^2dxdy.$$

Start by drawing the region. The desired integral is $\ds \int_{0}^{3} \int_{0}^{y}e^{y^2}dxdy.$ The integral can now be computed using first semester calculus tools (u substitution).

Start by drawing the region. It's a triangle, described (using vertical line segments) with $0\leq x\leq \sqrt{\pi}$ and $0\leq y\leq 2x$. We then compute $$\begin{align*} \int_0^{2\sqrt\pi}\int_{y/2}^{\sqrt{\pi}}\sin(x^2)dxdy &=\int_0^{\sqrt\pi}\int_{0}^{2x}\sin(x^2)dydx\\ &=\int_0^{\sqrt\pi}2x\sin(x^2)dx\\ &=\int_0^{\pi}\sin(u)du\\ &=-\cos(u)\bigg|_0^{\pi}\\ &=2 \end{align*} $$

The region can be described in polar coordinates with $0\leq \theta \leq \pi/4$ and $0\leq r\leq 2$. We know that in polar coordinates $x^2+y^2 = r^2$, and that $dA = rdrd\theta$, which means $$ ds\int_{0}^{\sqrt{2}}\int_{y}^{\sqrt{4-y^2}}e^{x^2+y^2}dxdy = \ds\int_{0}^{\pi/4}\int_{0}^{2}e^{r^2}rdrd\theta. $$

Recall that $dm = \delta dA$ and $dA = r dr d\theta$ in polar coordinates. Bounds for the region in polar coordinates are $0\leq \theta\leq 2\pi$ and $0\leq r\leq 2$. We can use Mathematica to verify these bounds are correct.

outerB = {theta, 0 , 2 Pi}
innerB = {r, 0, 3}
coordinates = {r Cos[theta], r Sin[theta]};
ParametricPlot[coordinates, outerB, innerB, Mesh -> {20, 0}]

The bounds above give the integral $$\begin{align} \bar x &= \frac{\int\int_R x dm}{\int\int_Rdm}\\ &= \frac{\int\int_R x (x+5)dA}{\int\int_R(x+5) dA}\\ &= \frac{\int_0^{2\pi}\int_{0}^{3} (r\cos\theta) (r\cos\theta+5)rdrd\theta} {\int_0^{2\pi}\int_{0}^{3} (r\cos\theta+5)rdrd\theta} . \end{align}$$ We can quickly compute this integral using Mathematica.

density = r Cos[theta] + 5;
mass = Integrate[density*r, outerB, innerB]
xbar = Integrate[r Cos[theta]*density*r, outerB, innerB]/mass
ybar = Integrate[r Sin[theta]*density*r, outerB, innerB]/mass
{xbar, ybar} // N

We can instead use Cartesian coordinates to set up the integral, which gives $$\begin{align} \bar x &= \frac{\int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} x(x+5)dydx} {\int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} (x+5)dydx}\\ &= \frac{\int_{-3}^{3}\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} x(x+5)dxdy} {\int_{-3}^{3}\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} (x+5)dxdy} . \end{align}$$ The code below draws the region, and computes the integrals, using Cartesian coordinates.

outerB = {x, -3, 3}
innerB = {y, -Sqrt[9 - x^2], Sqrt[9 - x^2]}
coordinates = {x, y};
ParametricPlot[coordinates, outerB, innerB, Mesh -> {10, 0}]
density = x + 5;
mass = Integrate[density, outerB, innerB]
xbar = Integrate[x*density, outerB, innerB]/mass
ybar = Integrate[y*density, outerB, innerB]/mass
{xbar, ybar} // N

Recall that a little length is given by $ds = v(t) dt$ (speed multiplied by a little time). Recall that the velocity $\vec v(t) = (3,3t^2)$ gives the speed $v(t) = \sqrt{9+9t^4}$. We then have $$\begin{align} \bar y &= \frac{\int_C y ds}{\int_C ds}\\ &= \frac{\int_{-1}^{2} t^3 \sqrt{9+9t^4} dt}{\int_{-1}^{2}\sqrt{9+9t^4} dt} . \end{align}$$ Here are the computations done with Mathematica.

ClearAll[r]
r[t_] = {3 t, t^3}
ds = Norm[r'[t]]
y = r[t][[2]]
Integrate[y ds, {t, -1, 2}]/Integrate[ ds, {t, -1, 2}]
% // N
ParametricPlot[r[t], {t, -1, 2}]

We'll draw this together, and use Integration.nb to check our work. We'll also discuss how changing the inner bounds will affect the integral with a few examples, such as $z=7-x$ instead of $z=5$, and $z=x$ or $z=-x^2$ instead of $z=0$.

There is a command to plot regions in Mathematica. Sometimes the corner's don't quite come together nicely, but the code is very short.

R = ParametricRegion[{x, y, z}, {{x, 0, 3}, {y, 0, 9 - x^2}, {z, 0, 5}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]

We'll draw this together, and use Integration.nb to check our work.

R = ParametricRegion[{x, y, z}, {{z, 0, 1}, {x, 0, 1 - z}, {y, 0, Sqrt[1 - x^2]}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]