Surface Integral of a Vector Field

Region B is the bounded by

Paraboloid:

$z = y + x^2 + y^2 -1$

and

Plane

$z = y$

We'll integrate over both surfaces,

$\displaystyle \int_{\partial B} \vec{F} \cdot \vec{dS} = \int_{\partial B_1} \vec{F} \cdot \vec{dS} + \int_{\partial B_2} \vec{F} \cdot \vec{dS}$

where $\partial B_1$ denotes the surface of the paraboloid and $\partial B_2$ is the surface of the plane.

The paraboloid can parameterized in cylindrical coordinates as

$p = (r \cos t , r \sin t, r^2 + r \sin t - 1 )$

Since $z \le y$ , then $r^2 \le 1$

$\vec{F} = (- x, 0, y - z) = (- r \cos t, 0, 1 - r^2 )$

Now the area differential $\vec{dS} = \pm p_t \times p_r \hspace{4pt}dt \hspace{4pt} dr$

$p_t = (- r \sin t, r \cos t, r \cos t )$

$p_r = ( \cos t, \sin t, 2 r + \sin t )$

so that

$p_t \times p_r = ( 2 r^2 \cos t , 2 r^2 \sin t + r , - r )$

since $(-r)$ is negative , this the right direction for an outward normal.

$\displaystyle \int_{\partial B_1} \vec{F} \cdot dS = \int_{t = 0 }^{2 \pi} \int_{r = 0 }^{1} (- r \cos t, 0, 1 - r^2 ) \cdot ( 2 r^2 \cos t , 2 r^2 \sin t + r , - r ) \hspace{4pt}dr \hspace{4pt} dt$

$= \displaystyle \int_{t = 0 }^{2 \pi} \int_{r = 0 }^{1} - 2 r^3 \cos^2 t + r^3 - r dr dt = \int_{t = 0 }^{2 \pi} \int_{r = 0 }^{1} - r^3 \cos 2 t - r dr dt$

Flipping the order of integration by integrating with respect to $t$ first, the first term vanishes, and the integral becomes

$= \displaystyle {2 \pi} \int_{r = 0 }^{1} - r dr dt = { 2 \pi } (- \dfrac{1}{2} ) = - \pi$

For the second integral (over the plane z= y ) , the parameterization of the plane $z = y$ is by taking two mutually orthogonal unit vectors that are

normal to the normal of the plane which is $(0, 1, -1)$ , so we can choose $u_1 = (1, 0, 0)$ , and $u_2 = (0, 1, 1) / \sqrt{2}$

hence a point on the plane is $p = u (1, 0, 0) + v (0, 1, 1)/sqrt(2) = (u , v / \sqrt{2} , v / \sqrt{2} )$

Now we can impose the condition that $r^2 = x^2 + y^2 <= 1$ , so that $u^2 + 1/2 v^2 <= 1$ , which is a region bounded by an ellipse, so we'll take $u = s \cos t , v = \sqrt{2} s \sin t$

, so that $p = ( s \cos t, s \sin t, s \sin t )$ , where $t \in [0, 2 \pi ]$ , and $s \in [0, 1]$ .

Now , with this parameterizaton, $\vec{F} = ( - x, 0, y-z) = (- s \cos t, 0, 0 )$

Since the normal to the plane is along the vector $( 0, 1, -1)$ , then $\vec{dS}$ is perpendicular to $\vec{F}$ and hence the second integral is zero.

The final answer is therefore,

$\displaystyle \int_{\partial B} \vec{F} \cdot \vec{dS} = - \pi$

A final note, it is possible to solve this problem using the divergence theorem, which states that

$\displaystyle \int_{\partial B} \vec{F} \cdot \vec{dS} = \int_B \text{div} \vec{F} dV$

we have $\text{div} \vec{F} = \dfrac{\partial F_x }{\partial x} + \dfrac{\partial F_y}{\partial y} + \dfrac{\partial F_z}{\partial z} = (-1) + 0 + (-1) = -2$

Hence,

$\displaystyle \int_{\partial B} \vec{F} \cdot \vec{dS} = -2 V$

where $V$ is the volume of region $B$ , the volume can found as follows:

Recall that a point on the paraboloid surface is,

$p = (r \cos t , r \sin t, r^2 + r \sin t - 1 )$

For cylindrical coordinates, the jacobian is $| J | = r$ .

Therefore,

$V = \displaystyle \int_{t=0}^{2 \pi} \int_{r = 0 }^{1} (1 - r^2) r dr dt = 2 \pi ( \dfrac{1}{2} - \dfrac{1}{4} ) = \dfrac{\pi}{2}$

Hence,

$\displaystyle \int_{\partial B} \vec{F} \cdot \vec{dS} = \int_B \text{div} \vec{F} dV = -2 \left(\dfrac{\pi}{2} \right) = - \pi$