Professor BG (Calc question)

[Shiro]

I've been trying to solve this integral for the past 3 hours and am on the verge of madness, so I thought BG might be interested in flexing its integration muscles tonight.

The question is: Integrate the function f(x,y,z) = 6 x + 3 y over the solid given by the "slice" of an ice-cream cone in the first octant bounded by the planes x=0 and y = sqrt(29/3)x and contained in a sphere centered at the origin with radius 10 and a cone opening upwards from the origin with top radius 8.

From what I can gather this is a 1/4 of a whole cone (in the first octant) going up in the z axis "sliced" by the x and y bounds. So, converting to spherical coordinates, I'm finding ɸ to be the angle between the z axis and the length of the cone at the point the cone and sphere touch, which should be sin^-1(8/10). θ similarly, should be the angle between the bounds on the xy plane, giving me cos^-1(x/sqrt(29/3)x) or just cos^-1(1/sqrt(29/3)). ρ should just be the radius of the sphere given, so 10. Setting up the integral I get: 3ρ^3*sin^2(ɸ)*(2cos(θ)+sin(θ)) dρ dɸ dθ so I should have:

0 <= ρ <= 10
0 <= ɸ <= sin^-1(4/5)
0 <= θ <= cos^-1(1/sqrt(29/3))

Integrating gives: (3/4)ρ^4*(ɸ/2-(1/4)sin(2ɸ))*(2sin(θ)-cos(θ)), which evaluated for the bounds...

http://i57.photobucket.com/albums/g2...e/calcsuck.jpg

Can anyone see any mistakes I've made? I've gone through this forward and back and can't figure out why this isn't being accepted...

[Woozie]

This looks fun, but my sleeping pills are already kicking in. I'll work on this tomorrow when I wake up if it isn't solved by then. I don't know when this is due so I hope tomorrow morning isn't too late.

[oldoldman]

If this isn't solved by the time I wake up I'll give it a go.

[Colenzo]

the answer is: 3143255r42852tjgegsgmaegnaemgq39g

[Melchiah]

If you're trying to input this algebraic answer into some piece of web-based grading software, it's more than likely the format.

A lot of universities are going to online homework grading, especially math and science departments, and most of the software is shitty as can be and only checks for one specific answer in one specific form, and rejects all others even if they are mathematically valid.

[Woozie]

Edit: Ignore this entire post. I solved the wrong problem.

I'm using the math convention that theta is the angle with respect to the z axis and theta is the polar coordinates angle. Physicists typically have it in reverse, but since you're in a math class, I'm assuming you're doing it the mathematician way too, so there shouldn't be any confusion.

Okay, here's what I did.

We want a surface integral of f(x,y,z) = 6 x + 3 y over a surface that has multiple sides. So I integrated over the different surfaces one at a time.

A top view of the surface would look something like this:

Spoiler: show

http://i26.tinypic.com/nv5c14.jpg

The top is dome shaped. On the edges (the vertical line x=0 and the slanted line y=sqrt(29/3)), the upper boundaries are dome/spherical shaped as well. It's the shape you'd get if you baked a spherical pie and cut a slice out. My drawing is just a two dimensional impression of that.

So this top surface can be described by

http://latex.codecogs.com/gif.latex?\rho=10

http://latex.codecogs.com/gif.latex?...0sin^{-1}(4/5)

http://latex.codecogs.com/gif.latex?...0\frac{\pi}{2}

So the function becomes:

http://latex.codecogs.com/gif.latex?...hi%20sin\theta

Integrating over the surface gives:

Edit: Fixed a mistake. Integrating over the surface gives:

http://latex.codecogs.com/gif.latex?...\theta%20d\phi

since http://latex.codecogs.com/gif.latex?\rho%20=%2010

then we get

http://latex.codecogs.com/gif.latex?...\theta%20d\phi

Now we integrate over the surface corresponding to the plane x=0. When the cone intersects, we're going to get something that looks somewhat like the slice of a circle

Spoiler: show

http://i26.tinypic.com/nv5c14.jpg

The vertical axis there is Z. It kinda got cut off at the top, but what that says is "X=0 plane" The slice of the circle that's somewhat shaded in is the part we're integrating over. The dotted line is imaginary. Like Canada. So pretend it isn't there. Like Canada.

Again, it's just a matter of integrating the function over this surface. Our function is now:

http://latex.codecogs.com/gif.latex?...\rho%20sin\phi

But this time our variables are P and phi. If you don't know the surface integral formula by heart for these two variables (which I don't), I'll evaluate it. If you know it by heart, skip the spoiler.

Spoiler: show

In general, a surface integral is:

http://latex.codecogs.com/gif.latex?...\vec{r}_v|dudv

where

http://latex.codecogs.com/gif.latex?\vec{r}_n

Is the vector obtained by differentiating the parametric equation by the variable n, and the two bars denote the absolute value of the cross product.

So for spherical coordinates with p and phi as the variables:

http://latex.codecogs.com/gif.latex?...hi%20cos\theta

The x component of http://latex.codecogs.com/gif.latex?\vec{r}_\phi is

http://latex.codecogs.com/gif.latex?...\phi%20sin\phi (I just differentiated the x component with respect to phi. I do the same with the y and z component and that gives me the entire vector for the phi variable. I do the same for the p variable, then take the cross product between the two.

You can verify that this is the cross prodcut:

http://latex.codecogs.com/gif.latex?...{vmatrix}\bigr

Is the cross product between the two vectors that are the derivatives of the parametric equation. Evaluate this cross product and get

http://latex.codecogs.com/gif.latex?...athbf{\hat{k}}

The magnitude of this vector p. So the integral will be

http://latex.codecogs.com/gif.latex?...0d\phi%20d\rho

Of course, you could have come to this conclusion simply by noting that if theta is constant, the spherical coordinates become polar in the y,z plane.

So our integral is

http://latex.codecogs.com/gif.latex?...0d\phi%20d\rho

I have to head to work, but I'll type up the evaluation of the next two regions when I get home (I'm only going to be at work for maybe an hour or so today).

Edit: I missed the bus to work, so I'll just have to work double hours tomorrow. I need to head to the store. I'll finish when I get back.

Edit: I just noticed I posted the same picture twice by accident. I'm not going to bother fixing it since they weren't asking for the surface integral anyways.

[Shiro]

Wow Woozie thanks a bunch. That's some impressive work
I think though in the problem they're looking for the volume for the object. They aren't specific at all, but it mentions the solid, and we never really covered anything dealing with surface area in class (some how), so I'm thinking I need to take the integral of all 3 variables simultaniously and find the space enclosed by the surfaces, not the surface area itself. Or am I jumping ahead of what you're doing? Your work does follow closely to what I had written up so I feel better knowing I was on the right track. A few questions though:

On your bounds for ɸ, you go from tan^1 to sin^-1 between your diagram and the written bounds. Freudian slip or is there some trig witchcraft going on there I'm missing. On the same note I had cos^-1(1/sqrt(29/3)) which evaluates a little different than your tan function for θ. My trig is pretty rusty so I'm probably making a mistake there somewhere.

Btw, your expression formating:

[Woozie]

Maybe I read this wrong. I thought they were asking to integrate over the surface, but it seems that they want you to integrate over the volume? Let me reread this.

Edit: You're right, it does say to integrate over a solid, not a surface. My bad D:

[Woozie]

On your bounds for ɸ, you go from tan^1 to sin^-1 between your diagram and the written bounds. Freudian slip or is there some trig witchcraft going on there I'm missing. On the same note I had cos^-1(1/sqrt(29/3)) which evaluates a little different than your tan function for θ. My trig is pretty rusty so I'm probably making a mistake there somewhere.

Edit: You're right, I did mix up some formulas. What I have written down is correct. Phi goes from 0 to tan^-1(4/5) and theta goes from sin^-1(sqrt(29/3)) to pi/2


This is what I have above
http://latex.codecogs.com/gif.latex?...\theta%20d\phi

This is what it should be.

http://latex.codecogs.com/gif.latex?...\theta%20d\phi

But I guess it doesn't matter because this was supposed to be integrated over a volume, not a surface. I'll redo this problem in a bit.

Edit: I may not get back to this for a few hours. Sorry D:

[Julian]

Integrals hurt my head so much, I hope I never have to see stuff like that again

Mind if I ask what class exactly this is for <.<;

[Shiro]

Integrals hurt my head so much, I hope I never have to see stuff like that again

Mind if I ask what class exactly this is for <.<;

Calc III: Multivariable Calculus

I don't think Integrals are so bad once you get used to them. There's just this aura of, well, almost magic around them when you take a big nasty function with varying density or what have you that looks impossible to solve, "Then I just draw this squiggly line here in front of it and give end points and I get an exact answer". The fundamental theorem still doesn't register with me intuitively; I've seen proofs and I know it works, but there's just some nagging part of my brain that doesn't want to believe it can work as flawlessly as it does..

[Woozie]

I've been trying to solve this integral for the past 3 hours and am on the verge of madness, so I thought BG might be interested in flexing its integration muscles tonight.

The question is: Integrate the function f(x,y,z) = 6 x + 3 y over the solid given by the "slice" of an ice-cream cone in the first octant bounded by the planes x=0 and y = sqrt(29/3)x and contained in a sphere centered at the origin with radius 10 and a cone opening upwards from the origin with top radius 8.

From what I can gather this is a 1/4 of a whole cone (in the first octant) going up in the z axis "sliced" by the x and y bounds. So, converting to spherical coordinates, I'm finding ɸ to be the angle between the z axis and the length of the cone at the point the cone and sphere touch, which should be sin^-1(8/10). θ similarly, should be the angle between the bounds on the xy plane, giving me cos^-1(x/sqrt(29/3)x) or just cos^-1(1/sqrt(29/3)). ρ should just be the radius of the sphere given, so 10. Setting up the integral I get: 3ρ^3*sin^2(ɸ)*(2cos(θ)+sin(θ)) dρ dɸ dθ so I should have:

0 <= ρ <= 10
0 <= ɸ <= sin^-1(4/5)
0 <= θ <= cos^-1(1/sqrt(29/3))

Integrating gives: (3/4)ρ^4*(ɸ/2-(1/4)sin(2ɸ))*(2sin(θ)-cos(θ)), which evaluated for the bounds...

http://i57.photobucket.com/albums/g2...e/calcsuck.jpg

Can anyone see any mistakes I've made? I've gone through this forward and back and can't figure out why this isn't being accepted...

Here you have

0 <= θ <= cos^-1(1/sqrt(29/3))

Here's a diagram that shows how I got http://latex.codecogs.com/gif.latex?...0\frac{\pi}{2}

Spoiler: show

http://i28.tinypic.com/4uc8ie.jpg

The angle here starts at

http://latex.codecogs.com/gif.latex?...\frac{29}{3}})

and you can see that it ends at pi/2. By the way, it should say sqrt(29/3), not sqrt(29/3)x. My bad.

That may have been your only mistake. I'm going to look over it for a second.

[Woozie]

Okay, your boundaries are correct if you fix the cos^-1 thing and use the tan^-1 that I got.

So you should get a triple integral like this:

http://latex.codecogs.com/gif.latex?...\rho%20d\theta

I with the limits of integration being the boundaries. I have no clue how you got your boundaries. All I did was substitute the spherical coordinate expressions for x and y, and used the fact that in spherical coordinates

http://latex.codecogs.com/gif.latex?...\rho%20d\theta

[Shiro]

Agg, that was it. I had some silly trig errors on the xy plane. I was using the x axis as the one of the boundaries rather than the y. Thanks for all your help!

[Woozie]

Agg, that was it. I had some silly trig errors on the xy plane. I was using the x axis as the one of the boundaries rather than the y. Thanks for all your help!

lol, I did the same thing at first and couldn't figure out why my answers weren't making sense.

[Abandon]

Necro bump, and fuck you for getting mayad. Pure brilliance. Such a loss.

[Sho]

It's one of my favorite threads that he's answered personally. How someone could solve something so complicated quickly is amazing.

My mans was definitely something else you don't see every day and/or generation.

[Kohan]

Necro bump, and fuck you for getting mayad. Pure brilliance. Such a loss.

I'm all for it.

[Ddz]

RIP Woozie. Your brilliance in the realm of mathematics and physics will be missed, along with your helpful and humble demeanor. May your soul rest in peace.

[Sera]

This shit is like being at the MOMA, you guys. I seriously have no fucking idea what he was doing but i'm still all like :O