Let $\vec{F}=\left(\frac{x}{(x^2+y^2+z^2)^{\frac32}},\frac{y}{(x^2+y^2+z^2)^{\frac32}},\frac{z}{(x^2+y^2+z^2)^{\frac32}}\right)$ be a vector field.
Denote $A=\{(x,y,z)\in \mathbb{R}^3 : x^2+y^2\leq 3, z=4-x^2-y^2\}$.
Find $\int_A FdS$.
My solution:
Using cylinder parameterization:
$x=r\cos\theta, y=r\sin\theta, z=4-r^2$.
The parameterization is $\vec{T}(r,\theta) = (r\cos\theta,r\sin\theta ,4-r^2)$.
The normal is $\vec{n} = \vec{T}_r \times \vec{T}_\theta = (2r^2cos\theta , 2r^2sin\theta, r).$
Now,
$\int_0^{2\pi}\int_0^3 (\frac{r\cos\theta}{((r\cos\theta)^2+(r\sin\theta)^2+(4-r^2)^2)^{1.5}},\frac{r\sin\theta}{((r\cos\theta)^2+(r\sin\theta)^2+(4-r^2)^2)^{1.5}},\frac{(4-r^2)}{((r\cos\theta)^2+(r\sin\theta)^2+(4-r^2)^2)^{1.5}})\cdot (2r^2\cos\theta , 2r^2\sin\theta, r) drd\theta= \int_0^{2\pi}\int_0^3 \frac{r^3+4r}{(r^4-7r^2+16)^{1.5}}drd\theta$
The answer of this integral is not $\pi$ but the correct answer is $\pi$.
I can't get where I am wrong, any help please.
As pointed out in the comments, the upper bound of $\ r$ is incorrect.
The integrand is correct. Substituting $\ r^2 = t$, we get:
$$\ I = \int_0^{2\pi} \frac{1}{2} \cdot \int_0^{3} \frac{t+4}{(t^2-7t+16)^{\frac{3}{2}}} dt d\theta$$
Note that
$$\ t^2 - 7t + 16 = \big(t-\frac{7}{2}\big)^2 + \frac{15}{4}$$
and
$$\ t + 4 = \big(t-\frac{7}{2}\big) + \frac{15}{2}$$
Rewriting $\ I$, we get:
$$\ I = \int_0^{2\pi} \frac{1}{2} \cdot \int_0^3 \frac{t-\frac{7}{2} + \frac{15}{2}}{\big(\big(t-\frac{7}{2}\big)^2 + \frac{15}{4}\big)^{\frac{3}{2}}}dt d\theta $$
$$ = \int_0^{2\pi} \frac{1}{2} \cdot \int_0^3 \frac{t-\frac{7}{2}}{\big(\big(t-\frac{7}{2}\big)^2+\frac{15}{4}\big)^{\frac{3}{2}}}dt d\theta + \int_0^{2\pi} \frac{1}{2} \cdot \int_0^3 \frac{\frac{15}{2}}{\big(\big(t-\frac{7}{2}\big)^2 + \frac{15}{4}\big)^{\frac{3}{2}}}dt d\theta $$
For the first integral, substitute $\big(t-\frac{7}{2}\big)^2+\frac{15}{4} = v$. For the second, substitute $\big(t-\frac{7}{2}\big) = \frac{\sqrt15}{2}\cdot \tan\phi$ and proceed.
The integral then evaluates to $\pi$, as said before.