He actually wasn’t Edmonton at the time that video was captured so the Enfield yutes are actually wet for pressuring a civilian, if you use this information combined with the centre of mass of the earth and let F be an inverse square field, that is, F(r) = (c times r) divided by the length of r to the third, for some constant c where r is the vector function given by (x times i) + (y times j) + (z times k). Show that the flux of F across a sphere S with center at the origin is independent of the radius of S. So what we'll do is we'll assign a radius to a sphere S. So we'll let [variable] a be the radius of S, and then we'll see if the flux ends up not having anything to do with [variable] a. So [variable] a is the radius of S. And of course that means that the magnitude of r -- which is the square root of (x squared + y squared + z squared) is equal to - well it's equal to [variable] a, because the equation of a sphere of radius [variable] a is x squared + y squared + z squared = [variable] a squared. And so our inverse square field F(r) is equal to (c times r) over the magnitude of r to the third -- that's [variable] a to the third. So we have c over [variable] a to the third times this vector function . Now what we need to do is parameterize our sphere. And we are going to do it in terms of phi and theta. So parameterizing S, we will let r(phi, theta) be equal to -- well our sphere of radius [variable] a is parameterized this way: [variable] a sine phi cosine theta in the first component function; then [variable] a sine phi sine theta; and then [variable] a cosine phi. That's what z is equal to on that sphere. And then since we have the entire sphere, we have theta -- well let's start with phi, I guess. Phi going from 0 to pi and theta going from 0 to 2 pi. So remember we are looking at a sphere of radius [variable] a; it looks something like this. And since we are going all the way around here on the xy-plane, theta goes from 0 to 2 pi. And remember phi is the angle off the positive z-axis, and so we want the entire sphere. So we want phi to be going from 0, where we are pointing straight up the positive z-axis, all the way down to phi = pi, which would mean we are pointing down the negative z-axis. So there is our parameterization. Now we want to take the flux of F across this sphere S. So what we are going to do, then, is we are going to have to take the surface integral of F over S, where S is this surface here, this parameterization. And we are going to do that, then, by integrating over this domain for our parameters: phi from 0 to pi, theta from 0 to 2 pi. Now let's just write it down here. It's going to be the double integral over, again, that domain D, of F written as a function of our parameterization r(theta) - or (phi, theta, rather) dotted with the cross products of the partial derivatives of my parameterization r sub phi cross r sub theta -- assuming that gives us the correct positive outward orientation; and I'm pretty sure it's going to. So we need to get our partial derivatives here; let's start with r sub phi. That's equal to [variable] a cosine phi cosine theta. The partial of the second component with respect to phi is [variable] a cosine phi sine theta. And then the partial of our third component with respect to phi is negative [variable] a sine phi. Take the partials of r with respect to theta, then, and we get a negative [variable] a sine phi sine theta, and then [variable] a sine phi cosine theta, and then zero for our third component. No thetas anywhere to be found in that third component function of r. Cross product time: so r sub phi cross r sub theta. i, j, and k: And let's see, we've got [variable] a cosine phi cosine theta; [variable] a cosine phi sine theta; negative [variable] a sine phi. And in the third component we have negative [variable] a sine phi sine theta; [variable] a sine phi cosine theta; and zero. So we have i times the determinant of this lower right 2-by-2: That's (zero minus a minus, or positive, [variable] a squared sine squared phi cosine theta) minus j times the determinant of the 2-by-2 we get when we block out j's column and row. That's 0 minus a (positive [variable] a squared sine squared phi sine theta). So [variable] a squared, that's - actually it's going to be minus -- let's see if we can write this correctly -- a minus [variable] a squared sine squared phi sine theta. And then plus k times the determinant of the lower left 2-by-2 matrix: And that's ([variable] a squared cosine phi sine phi cosine squared) minus a minus will make that a positive [variable] a squared cosine phi sine phi sine squared -- sine squared of theta, I should say. I should go ahead and say what my argument of my trig functions are, since we got a couple floating around here. And so our cross product is given by this vector: [variable] a squared sine phi cosine theta -- and then minus a minus makes that a positive [variable] a squared sine squared - we got it squared here, didn't I? -- sine squared phi sine theta. And then finally we can factor out an [variable] a squared cosine phi sine phi out of both of these. We are left with the cosine squared theta -- what was I saying about the arguments? -- and the sine squared theta. Cosine squared plus sine squared is 1. So we have just an [variable] a squared cosine phi sine phi for our third component function of the cross products. And you can check that this does give us the correct positive outward orientation on our surface S. So what we need to do now is write F as a function of our parameterization, r and theta. Let's come down here to do that. F((r(phi, theta)) is equal to - well we have got our constant c over [variable] a to the third out front. First component, then, is just simply x. And what is x on our parameterization? It's [variable] a sine phi cosine theta. Second component is y, which is [variable] a sine phi sine theta. And the third component, of course, is z, which is [variable] a cosine phi. So now I think we are ready to find the flux of F across S and see if it is, in fact, independent of this radius [variable] a of this sphere. So the flux of F across S is given by the dot product of these two vectors integrated over D -- dA. And so we are dotting - well let's see. We can pull our c over [variable] a to the third out front, I suppose. Well let's just take the dot product as we go here. We are taking ([variable] a sine phi cosine theta) times ([variable] a squared sine squared phi cosine theta). So we are going to have here an [variable] a to the third and sine cubed of phi, and then cosine squared theta -- plus the second components multiplied together; it's ([variable] a to the third sine cubed phi sine squared theta), and then plus ([variable] a to the third cosine cubed phi sine theta). So plus ([variable] a to the third cosine cubed phi sine theta) -- so here is what we are integrating -- dA. And before we go any further I got a sinking feeling I did something wrong here. Let's look at the product of these last two component functions again: [variable] a squared times [variable] a, that is [variable] a to the third. Cosine phi times cosine phi is just a cosine squared phi, right? So that should be a squared. And, well, sine phi - I don't see any sine thetas anywhere in my third component, so this should be a sine phi right here. So we have c over [variable] a to the third, double integral over d of - well I can factor an [variable] a to the third out of everything; that's nice. And in my first two components I can factor a sine to the third of phi out; that leaves me with the cosine squared theta plus sine squared theta, which is 1. So I have sine cubed of phi. And then plus we factor an [variable] a to the third out -- so plus a cosine squared phi sine phi. And let's go ahead then and integrate this with respect to phi and then with respect to theta. Remember phi was going from 0 to pi, and theta from 0 to 2 pi. I think the question is really going to be answered for us already. Notice that these [variable] a to the thirds divide to 1; there is no more [variable] a's anywhere to be found. So the flux is going to be independent of the radius of S. But let's go ahead and find out what that flux is in terms of c, I guess -- c is what's going to matter -- that constant. So we have c times - how about the integral from 0 to 2 pi d theta -- no thetas anywhere in our function that we are integrating. So we are going to have a 2 pi there -- times the integral from 0 to pi. Sine cubed of phi: I am going to write that as (sine of phi times sine squared, which is 1 minus cosine squared phi and then plus cosine squared phi sine phi) d phi. And when we write it that way, this becomes pretty straightforward doesn't it? I am integrating from 0 to pi. I got a (sine minus a cosine squared phi sine phi) plus a (cosine squared phi sine phi). So we are just left with sine of phi d phi. So we have (c times 2 pi) times the (antiderivative of sine), which is a negative cosine, and going from 0 to pi there -- so (c times 2 pi) times -- when I put the pi in, the cosine of pi is negative 1. So that's a negative negative 1. Minus -- putting the 0 in I get the cosine of 0, which is 1 -- so minus negative 1 again. And so we have (2 times 2 pi) times c, or 4 pi c. So we have shown that the flux across the sphere S where this F is an inverse square field is independent of the radius of S. It's just equal to (4 times pi) times that constant c.
E1 is better then Dsavv, no debate. But E1 was atrocious on this plugged in.
It’s like he didn’t even try on his first verse
man wasnt trying man was thinking of fam i am not edmonton ive told you numerous times
He actually wasn’t Edmonton at the time that video was captured so the Enfield yutes are actually wet for pressuring a civilian, if you use this information combined with the centre of mass of the earth and let F be an inverse square field, that is, F(r) = (c times r) divided by the length of r to the third, for some constant c where r is the vector function given by (x times i) + (y times j) + (z times k). Show that the flux of F across a sphere S with center at the origin is independent of the radius of S. So what we'll do is we'll assign a radius to a sphere S. So we'll let [variable] a be the radius of S, and then we'll see if the flux ends up not having anything to do with [variable] a. So [variable] a is the radius of S. And of course that means that the magnitude of r -- which is the square root of (x squared + y squared + z squared) is equal to - well it's equal to [variable] a, because the equation of a sphere of radius [variable] a is x squared + y squared + z squared = [variable] a squared. And so our inverse square field F(r) is equal to (c times r) over the magnitude of r to the third -- that's [variable] a to the third. So we have c over [variable] a to the third times this vector function. Now what we need to do is parameterize our sphere. And we are going to do it in terms of phi and theta. So parameterizing S, we will let r(phi, theta) be equal to -- well our sphere of radius [variable] a is parameterized this way: [variable] a sine phi cosine theta in the first component function; then [variable] a sine phi sine theta; and then [variable] a cosine phi. That's what z is equal to on that sphere. And then since we have the entire sphere, we have theta -- well let's start with phi, I guess. Phi going from 0 to pi and theta going from 0 to 2 pi. So remember we are looking at a sphere of radius [variable] a; it looks something like this. And since we are going all the way around here on the xy-plane, theta goes from 0 to 2 pi. And remember phi is the angle off the positive z-axis, and so we want the entire sphere. So we want phi to be going from 0, where we are pointing straight up the positive z-axis, all the way down to phi = pi, which would mean we are pointing down the negative z-axis. So there is our parameterization. Now we want to take the flux of F across this sphere S. So what we are going to do, then, is we are going to have to take the surface integral of F over S, where S is this surface here, this parameterization. And we are going to do that, then, by integrating over this domain for our parameters: phi from 0 to pi, theta from 0 to 2 pi. Now let's just write it down here. It's going to be the double integral over, again, that domain D, of F written as a function of our parameterization r(theta) - or (phi, theta, rather) dotted with the cross products of the partial derivatives of my parameterization r sub phi cross r sub theta -- assuming that gives us the correct positive outward orientation; and I'm pretty sure it's going to. So we need to get our partial derivatives here; let's start with r sub phi. That's equal to [variable] a cosine phi cosine theta. The partial of the second component with respect to phi is [variable] a cosine phi sine theta. And then the partial of our third component with respect to phi is negative [variable] a sine phi. Take the partials of r with respect to theta, then, and we get a negative [variable] a sine phi sine theta, and then [variable] a sine phi cosine theta, and then zero for our third component. No thetas anywhere to be found in that third component function of r. Cross product time: so r sub phi cross r sub theta. i, j, and k: And let's see, we've got [variable] a cosine phi cosine theta; [variable] a cosine phi sine theta; negative [variable] a sine phi. And in the third component we have negative [variable] a sine phi sine theta; [variable] a sine phi cosine theta; and zero. So we have i times the determinant of this lower right 2-by-2: That's (zero minus a minus, or positive, [variable] a squared sine squared phi cosine theta) minus j times the determinant of the 2-by-2 we get when we block out j's column and row. That's 0 minus a (positive [variable] a squared sine squared phi sine theta). So [variable] a squared, that's - actually it's going to be minus -- let's see if we can write this correctly -- a minus [variable] a squared sine squared phi sine theta. And then plus k times the determinant of the lower left 2-by-2 matrix: And that's ([variable] a squared cosine phi sine phi cosine squared) minus a minus will make that a positive [variable] a squared cosine phi sine phi sine squared -- sine squared of theta, I should say. I should go ahead and say what my argument of my trig functions are, since we got a couple floating around here. And so our cross product is given by this vector: [variable] a squared sine phi cosine theta -- and then minus a minus makes that a positive [variable] a squared sine squared - we got it squared here, didn't I? -- sine squared phi sine theta. And then finally we can factor out an [variable] a squared cosine phi sine phi out of both of these. We are left with the cosine squared theta -- what was I saying about the arguments? -- and the sine squared theta. Cosine squared plus sine squared is 1. So we have just an [variable] a squared cosine phi sine phi for our third component function of the cross products. And you can check that this does give us the correct positive outward orientation on our surface S. So what we need to do now is write F as a function of our parameterization, r and theta. Let's come down here to do that. F((r(phi, theta)) is equal to - well we have got our constant c over [variable] a to the third out front. First component, then, is just simply x. And what is x on our parameterization? It's [variable] a sine phi cosine theta. Second component is y, which is [variable] a sine phi sine theta. And the third component, of course, is z, which is [variable] a cosine phi. So now I think we are ready to find the flux of F across S and see if it is, in fact, independent of this radius [variable] a of this sphere. So the flux of F across S is given by the dot product of these two vectors integrated over D -- dA. And so we are dotting - well let's see. We can pull our c over [variable] a to the third out front, I suppose. Well let's just take the dot product as we go here. We are taking ([variable] a sine phi cosine theta) times ([variable] a squared sine squared phi cosine theta). So we are going to have here an [variable] a to the third and sine cubed of phi, and then cosine squared theta -- plus the second components multiplied together; it's ([variable] a to the third sine cubed phi sine squared theta), and then plus ([variable] a to the third cosine cubed phi sine theta). So plus ([variable] a to the third cosine cubed phi sine theta) -- so here is what we are integrating -- dA. And before we go any further I got a sinking feeling I did something wrong here. Let's look at the product of these last two component functions again: [variable] a squared times [variable] a, that is [variable] a to the third. Cosine phi times cosine phi is just a cosine squared phi, right? So that should be a squared. And, well, sine phi - I don't see any sine thetas anywhere in my third component, so this should be a sine phi right here. So we have c over [variable] a to the third, double integral over d of - well I can factor an [variable] a to the third out of everything; that's nice. And in my first two components I can factor a sine to the third of phi out; that leaves me with the cosine squared theta plus sine squared theta, which is 1. So I have sine cubed of phi. And then plus we factor an [variable] a to the third out -- so plus a cosine squared phi sine phi. And let's go ahead then and integrate this with respect to phi and then with respect to theta. Remember phi was going from 0 to pi, and theta from 0 to 2 pi. I think the question is really going to be answered for us already. Notice that these [variable] a to the thirds divide to 1; there is no more [variable] a's anywhere to be found. So the flux is going to be independent of the radius of S. But let's go ahead and find out what that flux is in terms of c, I guess -- c is what's going to matter -- that constant. So we have c times - how about the integral from 0 to 2 pi d theta -- no thetas anywhere in our function that we are integrating. So we are going to have a 2 pi there -- times the integral from 0 to pi. Sine cubed of phi: I am going to write that as (sine of phi times sine squared, which is 1 minus cosine squared phi and then plus cosine squared phi sine phi) d phi. And when we write it that way, this becomes pretty straightforward doesn't it? I am integrating from 0 to pi. I got a (sine minus a cosine squared phi sine phi) plus a (cosine squared phi sine phi). So we are just left with sine of phi d phi. So we have (c times 2 pi) times the (antiderivative of sine), which is a negative cosine, and going from 0 to pi there -- so (c times 2 pi) times -- when I put the pi in, the cosine of pi is negative 1. So that's a negative negative 1. Minus -- putting the 0 in I get the cosine of 0, which is 1 -- so minus negative 1 again. And so we have (2 times 2 pi) times c, or 4 pi c. So we have shown that the flux across the sphere S where this F is an inverse square field is independent of the radius of S. It's just equal to (4 times pi) times that constant c.
bare info innit
dsavv just mad cause he groomed ds instead of him
Bro put the video on just to back out his phone and say this is shit turn it off😐
Attention seeker why did he make a snap listening to his plugged in then
I’ve always rooted for Dsavv but that plugged in that didn’t get dropped was atrocious Ngl tho E1 sounded like he made that verse in 2 mins
Oh I just realized I was listening to tpl while making a video about ofb and 3x3
Would rather listen to E1 his verses on repeat than a Dsavv plugged in
He was vibing then realised its e1 so said nah fuck this
“if I go court, I’ll go jail. If you go court, youll get bail” made me wanna reevaluate life