Regions of integration; bounds and triple integrals.

[NewtonianAlch]

Homework Statement

My first problem is with 2ia) and 2ib), I got the correct answer, although not happy with my understanding of it.

http://img826.imageshack.us/img826/1038/443pr.jpg

The Attempt at a Solution

(2ia and 2ib)

The region that's of concern is the upper part between y = x^2 and y = 4 isn't it? I thought that since y ≤ 4 and y ≥ x^2, if it were the lower part then the bounds would reverse.

Using that reasoning I used the new bounds for x as: [√y ≤ x ≤ 2] because I thought, x is not going to get any larger than 2, and it's lower limit at any given point in time is √y, when integrating that, it's obvious the answer is not going to be elementary, so I just edited the bounds: [0 ≤ x ≤ √y] which works out, the only problem with this is now the region would appear to be the lower half and not the upper half that I originally thought it was given the initial bounds. What's happening here?

(2ii)

For this one, I'm having trouble describing the region. It's a sphere with radius ≤ 1, and based on the cuts given, it appears to be some kind spherical wedge, or even a slice of pizza pie, if the bounds for z is not too big. Is that correct?

 

[cjc0117]

for 2i, refer to the picture I attached. The red region is the region we're concerned about, not the blue region. When you take √y≤x≤2 you are describing the blue region. When 0≤x≤√y you are describing the red region (0≤y≤4 in both cases).

[STRIKE]for 2ii, the region is a wedge with its tip at the origin, one triangular face in the yz-plane and another triangular face in the plane y=x. The two faces intersect along the z-axis. The lateral (not sure if this is the right word) surface and the top are rounded. I'm bad at drawing 3D images and I don't have software so maybe someone else can help you there.

EDIT: In other words, the region is an eighth of a rounded cone.[/STRIKE]

 

[LCKurtz]

Here's a picture of the other one:

 

[NewtonianAlch]

Thanks for the responses, in regards to the second part, it's pretty difficult to visualise it. I did not realize there was a conic section until you mentioned it, but now I see the z^2 = x^2 + y^2.

How would one determine the angle for the rise? In other words, not the theta angle between y and x, but the phi angle, or the elevation? I'm thinking if the radius is a maximum of 1, then the phi angle would be Pi/4, the same as the theta angle since the point is (1,1,1). But I'm nto sure.

 

[NewtonianAlch]

I see the reason for the bounds in the first one now too, it's a bit confusing at first.

 

[cjc0117]

my description of the region for the second part was completely wrong. Sorry. Refer to LCKurtz diagram.

EDIT: It's actually the portion of the solid sphere [itex]x^{2}+y^{2}+z^{2}≤1[/itex] between the xz-plane and the plane y=x in the first octant, also hollowed out by the cone [itex]z=\sqrt{x^{2}+y^{2}}[/itex]. Does that make any sense, lol?

EDIT2: I included a diagram of the projections of the solid region onto each of the coordinate planes, though all the projections look the same so it's not very helpful.

Anyways, another way to find the lower limit of [itex]∅[/itex], though your reasoning seems perfectly valid, is to plug the equation of the cone into the equation of the surface of the sphere, simplify, switch to spherical coordinates, and solve for [itex]∅[/itex]:

[itex]x^{2}+y^{2}+(\sqrt{x^{2}+y^{2}})^{2}=1[/itex]

[itex]2(x^{2}+y^{2})=1[/itex]

[itex]2((ρsin∅cosθ)^{2}+(ρsin∅sinθ)^{2})=1[/itex]

[itex]2ρ^{2}sin^{2}∅(cos^{2}θ+sin^{2}θ)=1[/itex]

[itex]2ρ^{2}sin^{2}∅=1[/itex]

[itex]∅=sin^{-1}(\sqrt{\frac{1}{2ρ^{2}}})[/itex]

The radius of the sphere is [itex]ρ=1[/itex] so...

[itex]∅=\frac{π}{4}[/itex]

It's kind of intuitive though.

Also It is easy to see from the projections I included, namely the xz- and yz-plane projections, that the upper limit is [itex]\frac{π}{2}[/itex].