Randy Root
New member
- Joined
- Nov 10, 2013
- Messages
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Hello,
Thanks to anyone reading this, I appreciate your time and any help you may be able to offer.
As the title states, I am working on converting limits of integration from rectangular to spherical.
I understand the integrand requires the extra terms (rho^2)sin(phi).
So I have no trouble with that, but all of the videos and examples I can find concerning this exercise are problems that are concerning shapes close to perfect spheres. Obviously these shapes are easier to change to spherical coordinates as you often have something like p^2=x^2+y^2+z^2 which makes it easier. I am having trouble with problems that aren't as easy in spherical coordinates. For example, I am working out of Ron Larson's Calculus 9th edition, and problem 13 in section 7 chapter 14 states:
Triple Integral of x dz dy dx, where x is from -2 to 2, y is -sqrt(4-x^2) to sqrt(4-x^2), z is x^2+y^2 to 4
I have no problem putting this into cylindrical coordinates and evaluating, the answer is 0.
Also, I know the integrand should be (rho^3)sin^2(phi)cos(theta) d(rho)d(phi)d(theta).
I can figure out that theta goes from 0 to 2pi, as it does in cylindrical coordinates, but that is about it. Even looking at the limits on Calc-chat I am unable to understand how they got to those answers.
Anyway, let me know if anymore information is needed, and I thank you for your time.
Randy Root
So I have figured this problem out. In case anyone comes across something similar this is the method I used for solving the equation:
By observation of the rectangular limits, the equations we are working with are the plane z=4 and the cone z=x^2+y^2.
Drawing this graph automatically leads to the conclusion the theta limits go from 0 to 2pi. Next, to find phi we look at the intersection of the plane and surface, which gives a triangle with side 1 equal to 4 (z length) and side 2 equal to 2 (x or y length). These sides are opposite and adjacent to the angle phi we are looking for, so we take the arctan of the opposite over adjacent or arctan(1/2) which is our phi. But after this point the surface we are striking with our rho changes so this indicates we will have 2 integrals to represent this volume. So phi 1 goes from 0 to arctan(1/2). Next, we need to find rho 1. Well the surface our rho is reaching at this point is z=4. We can easily convert this to rho*cos(phi)=4 or rho=4sec(phi).
So our rho 1 goes from 0 to 4sec(phi).
For our second integral theta again obviously goes from 0 to 2 pi. First off, we know that phi 2 is going to start at arctan(1/2) to maintain continuity of the shape. To find the end point of our phi 2 we simply realize that the volume will be fully encompassed when phi reaches the xy plane, at phi=pi/2. So now all we need is to find out rho 2. By substituting the rectangular formula z=x^2+y^2 to spherical coordinates we get rho*cos(phi)=rho^2sin^2(phi). I factored these out, which left me with a sin^2 and cos^2 of theta, which simplifies to one. Anyway, solving this equation for rho, we get rho=cot(phi)csc(phi), which is our final point for rho 2 (rho 2 begins at 0).
So the equation we get looks like this:
Triple integral of p^3sin^2(phi)cos(theta) d(rho)d(phi)d(theta) with rho 1 from 0 to 4sec(phi), phi 1 from 0 to arctan(1/2), theta 1 from 0 to 2pi, and rho 2 from 0 to cot(phi)csc(phi), phi 2 from arctan(1/2) to pi/2 and theta 2 from 0 to 2 pi again. Even though changing to different coordinate systems is used to make the integral easier to evaluate, it looks like it doesn't always. This integral is actually fairly easy to evaluate in cylindrical coordinates, but no matter how you evaluate it the grand total volume of the integral is an astounding ZERO!!!!!
Anyway, hope this might help someone who was stuck on this like I was. Good luck in Cal 3 everybody!
Thanks to anyone reading this, I appreciate your time and any help you may be able to offer.
As the title states, I am working on converting limits of integration from rectangular to spherical.
I understand the integrand requires the extra terms (rho^2)sin(phi).
So I have no trouble with that, but all of the videos and examples I can find concerning this exercise are problems that are concerning shapes close to perfect spheres. Obviously these shapes are easier to change to spherical coordinates as you often have something like p^2=x^2+y^2+z^2 which makes it easier. I am having trouble with problems that aren't as easy in spherical coordinates. For example, I am working out of Ron Larson's Calculus 9th edition, and problem 13 in section 7 chapter 14 states:
Triple Integral of x dz dy dx, where x is from -2 to 2, y is -sqrt(4-x^2) to sqrt(4-x^2), z is x^2+y^2 to 4
I have no problem putting this into cylindrical coordinates and evaluating, the answer is 0.
Also, I know the integrand should be (rho^3)sin^2(phi)cos(theta) d(rho)d(phi)d(theta).
I can figure out that theta goes from 0 to 2pi, as it does in cylindrical coordinates, but that is about it. Even looking at the limits on Calc-chat I am unable to understand how they got to those answers.
Anyway, let me know if anymore information is needed, and I thank you for your time.
Randy Root
So I have figured this problem out. In case anyone comes across something similar this is the method I used for solving the equation:
By observation of the rectangular limits, the equations we are working with are the plane z=4 and the cone z=x^2+y^2.
Drawing this graph automatically leads to the conclusion the theta limits go from 0 to 2pi. Next, to find phi we look at the intersection of the plane and surface, which gives a triangle with side 1 equal to 4 (z length) and side 2 equal to 2 (x or y length). These sides are opposite and adjacent to the angle phi we are looking for, so we take the arctan of the opposite over adjacent or arctan(1/2) which is our phi. But after this point the surface we are striking with our rho changes so this indicates we will have 2 integrals to represent this volume. So phi 1 goes from 0 to arctan(1/2). Next, we need to find rho 1. Well the surface our rho is reaching at this point is z=4. We can easily convert this to rho*cos(phi)=4 or rho=4sec(phi).
So our rho 1 goes from 0 to 4sec(phi).
For our second integral theta again obviously goes from 0 to 2 pi. First off, we know that phi 2 is going to start at arctan(1/2) to maintain continuity of the shape. To find the end point of our phi 2 we simply realize that the volume will be fully encompassed when phi reaches the xy plane, at phi=pi/2. So now all we need is to find out rho 2. By substituting the rectangular formula z=x^2+y^2 to spherical coordinates we get rho*cos(phi)=rho^2sin^2(phi). I factored these out, which left me with a sin^2 and cos^2 of theta, which simplifies to one. Anyway, solving this equation for rho, we get rho=cot(phi)csc(phi), which is our final point for rho 2 (rho 2 begins at 0).
So the equation we get looks like this:
Triple integral of p^3sin^2(phi)cos(theta) d(rho)d(phi)d(theta) with rho 1 from 0 to 4sec(phi), phi 1 from 0 to arctan(1/2), theta 1 from 0 to 2pi, and rho 2 from 0 to cot(phi)csc(phi), phi 2 from arctan(1/2) to pi/2 and theta 2 from 0 to 2 pi again. Even though changing to different coordinate systems is used to make the integral easier to evaluate, it looks like it doesn't always. This integral is actually fairly easy to evaluate in cylindrical coordinates, but no matter how you evaluate it the grand total volume of the integral is an astounding ZERO!!!!!
Anyway, hope this might help someone who was stuck on this like I was. Good luck in Cal 3 everybody!
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