- #1
Can Triple Integrals Be Solved in Multiple Ways?
- Thread starter rado5
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- Integral Triple integral
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Homework Help Overview
The discussion revolves around the evaluation of triple integrals, specifically exploring whether they can be solved in multiple ways. The context involves spherical coordinates and the geometric interpretation of spheres and their intersections.
Discussion Character
- Exploratory, Mathematical reasoning, Problem interpretation
Approaches and Questions Raised
- Participants discuss dividing the integral into parts based on the geometry of the problem, specifically the intersection of spheres. There are varying approaches to setting up the integrals in spherical coordinates, with some participants suggesting different ranges for the variables involved.
Discussion Status
Some participants have provided guidance on how to approach the problem by suggesting specific integral setups. There is acknowledgment of different methods being explored, with some participants indicating that they have found solutions using alternative approaches. However, there is no explicit consensus on a single method.
Contextual Notes
Participants note the importance of correctly identifying the ranges for the variables in the integrals, with some expressing uncertainty about their initial setups. There is also mention of a reference solution from a textbook that contrasts with some participants' calculations.
- #2
HallsofIvy
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x^2+ y^2+ z^2= 2az when x^2+ y^2+ z^2- 2az= 0. We can complete the square by adding a^2 to both sides: x^2+ y^2+ z^2- 2az+ a^2= x^2+ y^2+ (z- a)^2= a^2. That is a sphere with center at (0, 0, a) and radius a. x^2+ y^2+ z^2= a^2 is, of course, a sphere with center at (0 0, 0) and radius a. The two spheres intersect when (z- a)^2 - z^2= z^2- 2az+ a^2- z^2= a^2- 2az= 0 or z= a/2, in which case x^2+ y^2+ z^2= x^2+ y^2+ a^2/4= a^2 or x^2+ y^2= 3a^2/4. That is, the intersection is a circle in the z= a/2 plane with center at (0, 0, a/2) and radius a\sqrt{3}/2.
In terms of spherical coordinates, each point on that circle of intersection has \rho= a and \theta= arcsin(\sqrt{3}/2)= \pi/6
The integral, then, should be done in two parts: For the first, \theta goes from 0 to \pi/6 and, for each \theta, \rho goes to the "upper sphere", [math]x^2+ y^2+ z^2= a^2[/math], from 0 to a.
For the second part, \theta goes from \pi/6 to \pi/2 and, for each \theta, \rho goes from 0 to the "lower" sphere x^2+ y^2+ z^2= 2az which in polar coordinates is \rho^2= 2a\rho cos(\theta). That is, for each \theta, \rho goes from 0 to 2a cos(\theta). (Note that when \theta= \pi/6, cos(\theta)= 1/2 so that \rho= a.)
\phi goes from 0 to 2\pi for both integrals.
(In order to be consistent with the original post, I have used "engineering" notation of spherical coordinates in which \theta is the "co-latitude" and \phi is the "longitude" rather than "mathematics" notation in which the two Greek letters are reversed. I have used \rho rather than "r" because that is the convention in either system.)
In terms of spherical coordinates, each point on that circle of intersection has \rho= a and \theta= arcsin(\sqrt{3}/2)= \pi/6
The integral, then, should be done in two parts: For the first, \theta goes from 0 to \pi/6 and, for each \theta, \rho goes to the "upper sphere", [math]x^2+ y^2+ z^2= a^2[/math], from 0 to a.
For the second part, \theta goes from \pi/6 to \pi/2 and, for each \theta, \rho goes from 0 to the "lower" sphere x^2+ y^2+ z^2= 2az which in polar coordinates is \rho^2= 2a\rho cos(\theta). That is, for each \theta, \rho goes from 0 to 2a cos(\theta). (Note that when \theta= \pi/6, cos(\theta)= 1/2 so that \rho= a.)
\phi goes from 0 to 2\pi for both integrals.
(In order to be consistent with the original post, I have used "engineering" notation of spherical coordinates in which \theta is the "co-latitude" and \phi is the "longitude" rather than "mathematics" notation in which the two Greek letters are reversed. I have used \rho rather than "r" because that is the convention in either system.)
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- #3
Quinzio
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Divide the integral in 2 parts:
0< \varphi < \pi/3 where \rho = a
and
\pi/3< \varphi < \pi where \rho = a\:sen2\varphi
z = \rho cos\varphi
So you solve
\int_{0}^{2\pi} \int_{0}^{\pi/3} \int_{0}^{a} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta <br /> +<br /> \int_{0}^{2\pi} \int_{\pi/3}^{\pi} \int_{0}^{2asen2\varphi} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta <br />
0< \varphi < \pi/3 where \rho = a
and
\pi/3< \varphi < \pi where \rho = a\:sen2\varphi
z = \rho cos\varphi
So you solve
\int_{0}^{2\pi} \int_{0}^{\pi/3} \int_{0}^{a} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta <br /> +<br /> \int_{0}^{2\pi} \int_{\pi/3}^{\pi} \int_{0}^{2asen2\varphi} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta <br />
Last edited:
- #4
Quinzio
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Quinzio said:
Only the firt part, if I didn't make mistakes:
=\int_{0}^{2\pi} \int_{0}^{\pi/3} \int_{0}^{a} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta
=\int_{0}^{2\pi} \int_{0}^{\pi/3} \left [ \frac{\rho^5}{5} (cos\varphi)^2 sen \varphi \: \right ]_0^a \: d\varphi\:d\theta
=\frac{ a^5 }{5} \int_{0}^{2\pi} \int_{0}^{\pi/3}(cos\varphi)^2 sen \varphi \: \: d\varphi\:d\theta
=\frac{ a^5 }{5} \int_{0}^{2\pi} \left[ \frac{-(cos\varphi)^3}{3} \: \right]_0^{\pi/3} \: d\theta=\frac{ a^5 }{5} \int_{0}^{2\pi} \left[ -\frac{3\sqrt3}{24} +\frac{1}{3} \: \right] \: d\theta=\frac{ a^5 }{5} \left( \frac{8-3\sqrt3}{24} \right) \int_{0}^{2\pi}\: d\theta
=2\pi \frac{ a^5 }{5} \left( \frac{8-3\sqrt3}{24} \right)
Last edited:
- #5
Quinzio
- 557
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Other part
=\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \int_{0}^{2asen \varphi} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta
=\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \left [ \frac{\rho^5}{5} (cos\varphi)^2 sen \varphi \: \right ]_{0}^{2asen \varphi} d\rho \: d\varphi\:d\theta
=\frac{32a^5}{5}\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \left (cos\varphi)^2 ( sen \varphi)^6 \: d\varphi\:d\theta
=\frac{32a^5}{5}\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \left (( sen \varphi)^6-( sen \varphi)^8) \: d\varphi\:d\theta
Now have fun solving sine power eight :)
=\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \int_{0}^{2asen \varphi} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta
=\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \left [ \frac{\rho^5}{5} (cos\varphi)^2 sen \varphi \: \right ]_{0}^{2asen \varphi} d\rho \: d\varphi\:d\theta
=\frac{32a^5}{5}\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \left (cos\varphi)^2 ( sen \varphi)^6 \: d\varphi\:d\theta
=\frac{32a^5}{5}\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \left (( sen \varphi)^6-( sen \varphi)^8) \: d\varphi\:d\theta
Now have fun solving sine power eight :)
- #6
rado5
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Dear HallsofIvy,
Thank you very much for your help. Yes my solution was wrong. At the end of my book the answer is \frac{59 \pi a^{5}}{480} and when I solved it with your method I got the correct answer. I was actually wrong about the ranges!
Dear Quinzio,
I used HallsofIvy's method and I could easily solve it!
Thank you very much for your help. Yes my solution was wrong. At the end of my book the answer is \frac{59 \pi a^{5}}{480} and when I solved it with your method I got the correct answer. I was actually wrong about the ranges!
Dear Quinzio,
I used HallsofIvy's method and I could easily solve it!
- #7
- #8
Quinzio
- 557
- 1
The two ways are really the same.
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