Double or triple integral that equals 30

Click For Summary
SUMMARY

The discussion focuses on constructing complex double or triple integrals that equal 30, specifically through the selection of appropriate regions and functions. Participants suggest using geometric shapes such as circles or spheres, with areas or volumes set to 30. An example provided involves a single integral with the integrand x², where the fundamental theorem of calculus is applied to derive a function that can be normalized to 30. The process involves choosing a function, defining a region or volume, performing the integration, and adjusting the result to meet the target value.

PREREQUISITES
  • Understanding of double and triple integrals
  • Familiarity with the fundamental theorem of calculus
  • Knowledge of geometric shapes and their properties (e.g., area of a circle, volume of a sphere)
  • Ability to manipulate functions and variables in calculus
NEXT STEPS
  • Explore methods for setting up double integrals over circular regions
  • Learn about triple integrals in spherical coordinates
  • Investigate normalization techniques for integrals
  • Review examples of integrals that yield specific values
USEFUL FOR

Students of calculus, mathematics educators, and anyone interested in advanced integration techniques and applications in mathematical modeling.

Mandanesss
Messages
37
Reaction score
0
For my assignment I have to come up with a really complex double or triple integral that equals 30.
Would you mind giving me some ideas?
 
Physics news on Phys.org
How about picking some region in the plane (a circle perhaps) that has area 30 or a region in 3d (a sphere perhaps) that has volume 30 and set up the corresponding integral?
 
Set up the integrals as a function and solve accordingly, I will start an example for a single integral.

Choose a function, set that as the integrand, set one limit of the integral as another variable, and solve the new function to 30.

eg I choose an integrand x^2.

So I let f(t)=\int^t_0 x^2 dx

Apply the fundamental theorem of calculus we get
f(t) = \frac{t^3}{3}
So now we can let f(t)=30 and solve accordingly.

Do the same but for a double or triple integral.
 
1) Pick a function
2) Pick a region/volume
3) Integrate
4) Normalize to 30
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K