Integrating the Integrand: Solving a Double Integral

Click For Summary
SUMMARY

The discussion focuses on solving the double integral ##\int_{z=0}^5 \int_{x=0}^4 \Big( \frac{xz}{ \sqrt{16-x^2}} +x \Big)dxdz##. The key method for integrating the term ##\frac{xz}{ \sqrt{16-x^2}}## involves using the substitution method with ##u=16-x^2##. Participants emphasize the importance of determining the integrability of the function ##f(x,z)## over the specified intervals, particularly noting that the integrand is undefined at ##x=4##. Proper justification of integrability allows for the evaluation of the double integral in any order.

PREREQUISITES
  • Understanding of double integration techniques
  • Familiarity with substitution methods in calculus
  • Knowledge of integrability conditions for functions
  • Basic proficiency in evaluating definite integrals
NEXT STEPS
  • Study the substitution method in detail, focusing on integrals involving square roots
  • Learn about the conditions for integrability of functions over specified intervals
  • Explore the evaluation of double integrals in different orders
  • Practice solving similar double integrals with varying limits and integrands
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques, as well as educators seeking to enhance their teaching methods for double integrals.

Ekramul Towsif
Messages
4
Reaction score
0

Homework Statement


##\int_{z=0}^5 \int_{x=0}^4 \Big( \frac{xz}{ \sqrt{16-x^2}} +x \Big)dxdz##

Homework Equations


double integration

The Attempt at a Solution


how do i integrate the term ##\frac{xz}{ \sqrt{16-x^2}}## though i know that ##\int x \, dx = \frac{x^2}{2}##
pls help me thoroughly :(
 
Physics news on Phys.org
Ekramul Towsif said:
how do i integrate the term ##\frac{xz}{ \sqrt{16-x^2}}##
The method used to solve that integral was most likely already covered in your past teacher's speeches.
Use substitution method, in particular ##u=16-x^2##. From this compute ##du/dx## and the new integral limits.
 
The integrand ##f(x,z)## is undefined when ##x=4##.
You will have to discuss integrability of ## x \to f(x,z) ## on ##[0,4[##, and of ## z \to \int_0^4 f(x,z) \ dx ## on ## [0,5]##.
If you can justify this, your double integral is well-defined and you can evaluate the integrals in any order you like.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
Replies
4
Views
3K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
3
Views
2K