Been trying to answer this problem for hours. Here is the text of the problem:
Let $W$ be the solid bounded by the paraboloid $x = y^2 + z^2$ and the plane $x = 4$. Let $F = 5x\hat{i} + y\hat{j} + z\hat{k}$
(a) Let $S_1$ be the surface of the paraboloid oriented in the negative $x$ direction. Find the flux of the vector field through the surface $S_1$.
I am stuck with the integral over the paraboloid of $y^2 + z^2 – y – z$. If that is right, my problem is in trying to write a double integral of that in cylindrical coordinates. Can anybody help me?
Best Answer
Maybe I can take you part way.
You'll need the normal unit vector to the surface, which is found through the gradient. If $g(x, y, z) = y^2 + z^2 - x = 0$, then
$$\hat{n}(x, y, z) = \frac{\vec\nabla g(x, y, z)}{|\vec\nabla g(x, y, z)|} = \frac{-\hat{x} + 2y \hat{y} + 2z \hat{z}}{\sqrt{1+4y^2+4z^2}}.$$
Then the flux density is
$$\hat{n}\centerdot \vec{F} = \frac{-5x + 2y^2 + 2z^2}{\sqrt{1+4y^2+4z^2}}.$$
From here, you might notice that this is in terms of $y^2 + z^2 = r^2$ which is the distance from the $x$ axis:
$$\hat{n}\centerdot \vec{F} = \frac{-5x + 2r^2}{\sqrt{1+4r^2}}.$$