Q33 consider the solid bounded by the shapes described by...

Q33: Consider the solid bounded by the shapes described by $x^2+y^2=4$ and $y^2+z^2=4$. An eighth of the solid is shown below. Evaluate the volume $V$ of the solid. Q33: Find the flux of $F=-yi+xj+6z^2k$ out of the closed surface $S$ bounded by the paraboloids; $z=4-x^2-y^2$ and $z=x^2+y^2$. (Hint: You may consider using polar coordinates at some point)

Transcript

00:01 We are asked to calculate the flux of the given surface, and we are given that z is in forms of g of x, y, mainly that z is equal to x and y.

00:16 So we will have to use the equation that the surface of our flux integral is equal to negative p times the partial with respect to x, minus q times the partial with respect to y plus r, da.

00:40 And in this case, f is going to be equal to our equation.

00:44 So f is equal to p .i.

00:47 Plus qj plus rk.

00:53 And we are given that f is equal to x, y, i, plus y, z, j, plus y, j, plus z x k.

01:11 So since we're also given that z is equal to g of x y, which is equal to 4 minus x squared minus y squared, if we are starting to plug into our formula, we actually already have everything we need besides the x partial and the y partial.

01:33 What the x partial of z is going to be dg, dx, which is equal to negative x squared derived, which is negative 2x.

01:44 And it will be the same for y.

01:46 So dg, d, d, y is equal to negative 2y.

01:52 So now we have both the partials.

01:54 We have p, q, and r, because pq and r are the terms attached to i, j, and k.

02:00 And that is enough to plug in for our integral.

02:04 Because we are also given that x goes from 0 to 1, and y goes from 0 to 1.

02:14 So, plugging into our integral, we'll have the integral from 0 to 1, integral from 0 to 1, of negative xy, as that's negative p, multiplied by negative 2x minus y z, multiplied by negative 2y, plus z x, d .a.

02:42 And this will be equal to the integral from 0 to 1, integral from 0 to 1, of 2x2.

02:49 Squared y plus 2 y squared x x plus z x d y d x and this particular problem the order really won't change anything d x d y would also work just as well as d y d x and now substituting out z because we know z is equal to 4 minus x squared minus y we need to get rid of the z in order to make this a integral in terms of only x and y so doing so we'll give the integral from 0 to 1 of 2x squared y plus 2 y squared multiplied by 4 minus x squared minus y squared plus x multiplied by 4 minus x squared minus y squared of dy t x and this will be equal to the same integral 0 to 1 0 to 1 of 2 x squared y plus 8 y squared 1 minus 2x squared, y squared.

03:59 This isn't going to be extremely pretty, by the way.

04:02 This is two three -term foils.

04:05 Minus 2x squared to y squared, minus 2y to the fourth, plus 4x minus x cubed, x cubed, minus x, y squared, d, y, squared, d, d, x.

04:21 And now we are good to do the first integral with respect to y...

Question

Please give Ace some feedback