Where did I go wrong in my setup for this triple integral problem?

[Ronaldo95163]

Hey guys I've been working some triple integration problems and I've stumbled across a question that I'm having problems with
So from the picture below my solution is incorrect and I can't seem to figure out where I went wrong. Is my setup for the integrals correct or is that where I've made my mistake.


1. Homework Statement
Q15 in the posted image
(Further Problems 15 pg 658 in the solutions posted)

Homework Equations

The Attempt at a Solution

So I found the intersection of the two surfaces z = sqrt(x^2+y^2) and z^2 = 4(x^2+y^2) and found this to be z=2a.
On the y-z plane the solid basically leaves a region bounded by z=2|y| and the sphere. Since z=2a then y = a. I used this to calculate phi as arctan(a/2a) and rho as 2asec(phi).
In the x-y plane is basically a circle of radius a...I used this to get my range for theta which was 0 to 2π

Then I setup the triple integral as follows(With 1 as the integrand because it's a Volume calculation for the solid formed)

My Right most integral for rho goes between 0 and 2asec(phi)
Middle goes from 0 to arctan(0.5)
Left most goes from 0 to 2 pi

My answer was 2a^3/π

 

[BvU]

Can't read it. Please type what you want us to help you with.
And read the guidelines.

 

[Ronaldo95163]

^Sorry About that...I've fixed it

 

[BvU]

How hard is it to type the problem statement ? Anyway:

In the x-y plane is basically a circle of radius a

In the x-y plane $z=0$ so x and y are 0 too. A circle of radius 0, therefore...

 

[Ronaldo95163]

lol I thought it would have been easier to understand straight from the textbook.
But it said z>=0 so at 0 wouldn't the solid be lying on the x-y plane?
Because if I substitute z= 0 into the equation of the sphere then you get a circle

 

[SammyS]

lol I thought it would have been easier to understand straight from the textbook.
But it said z>=0 so at 0 wouldn't the solid be lying on the x-y plane?
Because if I substitute z= 0 into the equation of the sphere then you get a circle

It is possible to post an image directly in a post like ...

I suppose the following is the textbook solution.

You need to determine the intersection of the cone and sphere, not the intersection of the sphere and x-y plane.

 

[Ronaldo95163]

Hmm okay so I solved for z already which I got as 2a ---> Resulting in y = a. Substituting those values back into the equation for the cone I get x = 0.
So the curves intersect in the yz plane?

But if you substitute z=2a in that same eqn for the cone you get a circle of radius a. So the region of intersection is a circle that's projected on the xy plane?

 

[SammyS]

Hmm okay so I solved for z already which I got as 2a ---> Resulting in y = a. Substituting those values back into the equation for the cone I get x = 0.

How does that give y = a?

But if you substitute z=2a in that same eqn for the cone you get a circle of radius a. So the region of intersection is a circle that's projected on the xy plane?

The intersection of those two surfaces is at z = 2a. The intersection is indeed a circle.

 

[Ronaldo95163]

Well I decided to use spherical coordinates from the start and normally to get my bounds for rho and phi I set x=0 to get the projection on the yz plane. So from the cone's equation I ended up with z=2y and I substituted z=2a into that which gave y=a.

 

[SammyS]

Well I decided to use spherical coordinates from the start and normally to get my bounds for rho and phi I set x=0 to get the projection on the yz plane. So from the cone's equation I ended up with z=2y and I substituted z=2a into that which gave y=a.

y = a if you set x = 0 and z = 2a. More generally, y = ±a for this case

Also, x = ±a, if you set y = 0 and z = 2a.

 

[Ronaldo95163]

Ohh I see.
From there I graphed the region onto the yz plane.
So its the graph z= 2|y| intersected by the sphere. I then used a bit of trigs to get that cos(phi) = 2a/rho giving rho as 2asec(phi) and tan(phi) as 1/2 -> phi = arctan(1/2)So my bounds were:

0<rho<2asec(phi)
0<phi<arctan(1/2)

 

[LCKurtz]

Ohh I see.
From there I graphed the region onto the yz plane.
So its the graph z= 2|y| intersected by the sphere. I then used a bit of trigs to get that cos(phi) = 2a/rho giving rho as 2asec(phi) and tan(phi) as 1/2 -> phi = arctan(1/2)So my bounds were:

0<rho<2asec(phi)
0<phi<arctan(1/2)

The equation of the sphere in spherical coordinates is not $\rho = 2a\sec \phi$. So your "bit of trigs" is wrong. Show us how you get $\rho$ for the sphere by expressing its equation in spherical coordinates.

As an aside, since the equation of the sphere isn't particularly simple in spherical coordinates (as you will see when you work it out), this problem is actually easier to set up in cylindrical coordinates.

 

[Ronaldo95163]

This was the diagram I used to get phi from the yz plane.

I just did cos(phi) which is 2a/rho and found rho from that to get 2asec(phi). But I understand where you're coming from...if I change the eqns I get z = 2r and z=sqrt(2a^2-r)+a.

But for spherical...I can't use the diagram like that to get rho since its the distance from the origin to a point on the sphere?

EDIT: I think i saw where I went wrong. I only set x of the cone to 0.
Setting x=0 in the cylinder gives y(sq)+(z-a)(sq) = 2a(sq)
expanding and simplifying gives y^2+z^2-2az = a^2.

From the line z=2y, I can make a substitution into that and solve for either y or z but that would get messy...I guess this is why I should have used cylindrical coordinates then?

 

[BvU]

i hope you switched the z and the y in the picture by accident ?

 

[Ronaldo95163]

Woa haha yeah I did sorry about that

 

[BvU]

Now: is there any difference with the corresponding picture of the x-z plane ? Or, for that matter, with the picture of the r-z plane ($\ r^2=x^2+y^2\$) ?

 

[LCKurtz]

The equation of the sphere in spherical coordinates is not $\rho = 2a\sec \phi$. So your "bit of trigs" is wrong. Show us how you get $\rho$ for the sphere by expressing its equation in spherical coordinates.

You haven't responded to this. You have $x^2+y^2+(z-a)^2 = 2a^2$ given as the equation of the sphere. You don't need any geometry to express that in spherical coordinates. Expand it out and use the appropriate equations to change $x^2+y^2+z^2$ and $z$ to spherical coordinates. Then solve the resulting equation for $\rho$. Do it and show your equations here. Getting the proper equation for $\rho$ is all you have left to do to set it up in spherical coordinates.

 

[Ronaldo95163]

Sorry about that...didn't get to respond because of assignments etc for other courses.
But I worked that out already but I didnt post it so here it is:

x(sq)+y(sq) + z(sq)-2az+a(sq) = 2a(sq)
rho(sq)-2az = a(sq)
rho(sq)-2a(rho)(cos(phi)) = a(sq)
rho(sq)-2a(rho)(cos(phi)) -a(sq) = 0

Using the quadratic formula I got rho = acos(phi)+/-asin(phi)

 

[SammyS]

Sorry about that...didn't get to respond because of assignments etc for other courses.
But I worked that out already but I didnt post it so here it is:

x(sq)+y(sq) + z(sq)-2az+a(sq) = 2a(sq)
rho(sq)-2az = a(sq)
rho(sq)-2a(rho)(cos(phi)) = a(sq)
rho(sq)-2a(rho)(cos(phi)) -a(sq) = 0

Using the quadratic formula I got rho = acos(phi)+/-asin(phi)

Those look like arcsine and arccosine functions.

Probably better as rho = (a)cos(phi) ± (a)sin(phi) .

However, I'm pretty sure you made a mistake in applying the quadratic formula.

 

[Ronaldo95163]

oops made a sign error

Im getting cos(phi) +/- sqrt(1+cos^2(x) )

And btw is it wrong to use geometry to get rho and phi in certain cases?

For example this qn(Not switching the topic of the thread btw...just found a question where I think geometry may be applicable)

 

[BvU]

Integral of z/r³ is indeed off topic, but the integral of 1 is interesting (and pretty easy to do). The remainder of your integral is then also a lot less complicated than where you are heading with your current change of variables ...

 

[LCKurtz]

Sorry about that...didn't get to respond because of assignments etc for other courses.
But I worked that out already but I didnt post it so here it is:

x(sq)+y(sq) + z(sq)-2az+a(sq) = 2a(sq)
rho(sq)-2az = a(sq)
rho(sq)-2a(rho)(cos(phi)) = a(sq)
rho(sq)-2a(rho)(cos(phi)) -a(sq) = 0

That's $\rho^2 -2a\rho\cos\phi - a^2 = 0$ and it's correct. It's just as easy to type in Latex and you should look over the examples in https://www.physicsforums.com/help/latexhelp/

Using the quadratic formula I got rho = acos(phi)+/-asin(phi)

That isn't correct. Fix it and then use Latex to type your set up triple integral to show us. You may find it difficult to integrate but getting it set up correctly is good learning practice. You only have that one small step left to do that. Then, as I told you before, you should set it up in cylindrical coordinates, where you will find the integral easy to do.