One can generate a torus as follows: $\vec{g}=((b+a\cos u)\cos v, (b+a\cos u)\sin v, a \sin u)$. To find its area, we can use a surface integral of the form $S=\iint_{D_{uv}} {\lVert \frac{∂g}{∂u} \times \frac{∂g}{∂v} \rVert \, du \, dv}$. However, in the case of torus the integral becomes seemingly unnecessarily tedious to evaluate. Are there nicer approaches to evaluating the surface area using a surface integral?
Integration – How to Calculate the Surface Area of a Torus
areaintegrationnormed-spaces
Related Solutions
You actually had just about everything right, except that you skipped an important step: your normal vector to the surface $ \ \vec{ds} \ $ is correct, but you need to integrate its length over the surface of the cone nappe in order to obtain the surface area.
I'll generalize the problem a little, since the choice of proportions for the cone hides one of the factors in the surface area result. For a cone nappe with a height $ \ h \ $ and a "base radius" $ \ r \ $ , we can use similar triangles to find the parametrization (using your notation)
$$ x \ = \ \left( \frac{r}{h} \right) u \ \cos \ p \ \ , \ \ y \ = \ \left( \frac{r}{h} \right) u \ \sin \ p \ \ , \ \ z \ = \ u \ \ , $$
with the domain $ \ 0 \ \le \ u \ \le \ h \ , \ 0 \ \le \ p \ < \ 2 \pi \ $ . An "upward" normal vector is then given by
$$ \vec{R_u} \ \times \ \vec{R_p} \ \ " = " \ \ \left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ \left( \frac{r}{h} \right) \cos \ p&\left( \frac{r}{h} \right) \sin \ p\quad&1\\ -\left( \frac{r}{h} \right) u \ \sin \ p&\left( \frac{r}{h} \right) u \ \cos \ p\quad&0\end{array}\right| $$
$$ = \ \langle \ -\left( \frac{r}{h} \right) u \ \cos \ p \ \ , \ \ -\left( \frac{r}{h} \right) u \ \sin \ p \ \ , \ \ \left( \frac{r}{h} \right)^2 u \ \rangle \ \ . $$
So, up to this point, your procedure is fine. What is needed now is the "norm" of this vector:
$$ \| \ \vec{R_u} \ \times \ \vec{R_p} \ \| \ \ = \ \ \left[ \ \left( \frac{r}{h} \right)^2 u^2 \ \cos^2 \ p \ + \ \left( \frac{r}{h} \right)^2 u^2 \ \sin^2 \ p \ + \ \left( \frac{r}{h} \right)^4 u^2 \ \right]^{1/2} \ \ . $$
$$ = \ \ \left[ \ \left( \frac{r}{h} \right)^2 u^2 \ + \ \left( \frac{r}{h} \right)^4 u^2 \ \right]^{1/2} \ = \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \ u \ \ . $$
It is the "magnitude" of the infinitesimal patches associated with the normal vectors that we wish to integrate over the domain of the parameters. Thus,
$$ S \ \ = \ \ \int_0^{2 \pi} \int_0^h \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \ u \ \ du \ dp $$
$$ = \ \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \int_0^{2 \pi} dp \ \int_0^h \ u \ \ du $$
$$ = \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \cdot \ 2 \pi \ \cdot \ \left(\frac{1}{2}u^2 \right) \vert_0^h \ \ = \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \cdot \ \pi \ h^2 $$
$$ = \ \pi \ h \ \cdot \ \left(\frac{r}{h} \right) \ \cdot \ h \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ = \ \pi \ r \ \sqrt{ r^2 \ + \ h^2 } \ \ , $$
or $ \ \pi \ $ times the "base radius" times the "slant height" of the cone nappe, as the surface area is frequently expressed. In your use of the "standard cone", for which $ \ r \ = \ h \ $ , this formula gives us $ \ S \ = \ \pi \ \sqrt{2} \ h^2 \ $ , as you will find for your calculations, with the restoration of the omitted step.
For $\iint_S\langle3x,-z,y\rangle\cdot d\vec{n}$, with orientation toward the origin, I get $$\begin{align} &\iint_S\langle3x,-z,y\rangle\cdot d\vec{n}\\ &=\int_0^{\pi/2}\int_0^{\pi/2}\langle12\cos(u)\sin(v),-4\cos(v),4\sin(u)\sin(v)\rangle\cdot-\langle\cos(u)\sin(v),\sin(u)\sin(v),\cos(v)\rangle\,16\sin(v)\,du\,dv\\ &=-192\int_0^{\pi/2}\int_0^{\pi/2}\cos^2(u)\sin^3(v)\,du\,dv\\ &=-192\int_0^{\pi/2}\cos^2(u)\,du\int_0^{\pi/2}\sin^3(v)\,dv\\ &=-192\cdot\frac{\pi}{4}\cdot\frac{2}{3}\\ &=-32\pi \end{align}$$
which matches your answer. Sometimes there is a bug in the problem code for these software platforms. Is this MyMathLab? WeBWorK? WebAssign? Something else?
Best Answer
Another way to proceed would be by writing out the surface integral using differential forms. To this end we need to set up a chart and a coordinate system on the Torus. Fortunately we need only one chart and the coordinate system can be made global (using the coordinates already introduced in the question $u,v$), the surface integral will be
$$ S=\int_{M} ab~ du \wedge dv $$ The integration runs over the coordinate range of the open cover $M=T^2$, that is $u \in [0, 2\pi], ~v \in [0, 2\pi]$ leading to
$$ S= ab \int_0^{2\pi}du \int_0^{2\pi} dv = 4\pi^2 ab $$