A uniform fluid flowing vertically downward (heavy rain) is described by the vector field $F=(0,0,-1)$. Find the total flow through the cone $z=\sqrt{x^2+y^2}$, $x^2+y^2 \leq 1$.
b)Now consider $F=(-\frac{\sqrt{2}}{2},0,\frac{-\sqrt{2}}{2})$ and Find the total flow through the cone $z=\sqrt{x^2+y^2}$, $x^2+y^2 \leq 1$
Attempt a)
We shall use the expresion
$$\int {F dS}=\int \int_{D}F \cdot ( T_{u} \times T_{v})=\int \int_D (-Pg_x-Qg_y+R)dA$$
Notice that the region $S$ alredy was given $g(x,y)=z(x,y)=\sqrt{x^2+y^2}$
$$g_x=\frac{x}{\sqrt{x^2+y^2}}$$
$$g_y=\frac{y}{\sqrt{x^2+y^2}}$$
doing the substitution on the integral
$$\iint_{D}(-0g_x-0g_y+(-1))dA$$
Since $x^2+y^2 \leq 1$ we shall use polar cordinates
$$x=r \cos \theta $$
$$y=r \sin \theta$$
where $r\in [0,1]$ and $\theta \in [0,2\pi]$
Finally doing the integration
$$ \int_{0}^{2 \pi} \int_{0}^{1}-rdrd \theta=-\pi $$
Now for $b)
We shall use the expresion
$$\int {F dS}=\int \int_{D}F \cdot ( T_{u} \times T_{v})=\int \int_D (-Pg_x-Qg_y+R)dA$$
Notice that the region $S$ alredy was given $g(x,y)=z(x,y)=\sqrt{x^2+y^2}$
$$g_x=\frac{x}{\sqrt{x^2+y^2}}$$
$$g_y=\frac{y}{\sqrt{x^2+y^2}}$$
doing the substitution on the integral
$$\iint_{D}(\frac{\sqrt{2}}{2}g_x-0g_y+(\frac{-\sqrt{2}}{2}))dA$$
$$\frac{\sqrt{2}}{2} \iint_{D}(\frac{x}{\sqrt{x^2+y^2}}-1)dA$$
Usinng polar coordinates we get
$$\int_{0}^{2 \pi} \int_{0}^{1} (cos \theta-1)r dr d\theta=-\frac{\pi}{\sqrt{2}}$$
Are my answers right or probably i do a mistake applying the formula.
Best Answer
Please note your normal vector is pointing upward. As per question, You need normal vector pointing downward through the cone surface. So the signs of your final answer will change otherwise it looks fine.
From cross product, you get $(-\frac{x}{\sqrt{x^2+y^2}}, -\frac{y}{\sqrt{x^2+y^2}}, 1)$.
Please see the third coordinate is positive that gives us upward direction but given we want downward direction, you should take the normal vector $(\frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}}, -1)$.
Now please do the same calculations and you will get opposite sign in both your answers.
By the way, instead of converting to polar coordinates at the time of integration, you could also parametrize the cone as below -
$T(r, \theta) = (r \cos \theta, r \sin \theta, r) \,$ $(0 \leq r \leq 1, 0 \leq \theta \leq 2\pi)$
So you get $T_{\theta} \times T_{r} = \, (r \cos \theta, r \sin \theta, -r)$
Integral becomes $\iint_D F.(T_{\theta} \times T_{r}) \, dr \, d\theta$