Integration Help: Solving Last Problem in Multivariable Calculus

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In summary, the problem discussed is a change in variables problem in multivariable calculus, where the integral 4/3 times the integral from 0 to 1 of (1-u^2)^(1/2)*(2u^2+1)du is given. The conversation includes an attempt at using substitution to solve the problem, but it is deemed unsuccessful. The suggested method is to split the integrand into smaller pieces and use a table of integrals to find the general solution. The transformed integral is then shown, and the conversation continues with a clarification on a missed cosine term in the transformation.
  • #1
mkkrnfoo85
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I was doing a change in variables problem in multivariable calculus and I got stuck on the last integration.

[tex]\frac {4}{3} \int_{u=0}^{1} (1-u^2)^{\frac {1}{2}}(2u^2+1)du[/tex]

I don't think substitution works. Can anyone show me an easy way to solve this? Thanks.
 
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  • #2
Typically [itex] u=\sin t [/itex]

Daniel.
 
  • #3
For problems like this in general, you should try spliting up any integrand into as many pieces as possible...

Here you can rewrite the integrand obviously as 2u^2[(1-u^2)^0.5] + (1-u^2)...

and being the lazy engineering student that I am... I would look these up in a table of integrals--- which would definitely have the general solutions.
 
  • #4
Don't listen to an engineering student giving advice in anything but engineering...:wink:


The transformed integral should be

[tex] \frac{4}{3}\int_{0}^{\frac{\pi}{2}} \cos^{2}t\left(2\sin^{2}t+1\right) dt [/tex]

Use the double angle formulas to get it simplified.

Daniel.
 
  • #5
did u miss [tex](1-u)^{1/2}[/tex] -> [tex] \int cos t (2 sin^2 t + 1) dt[/tex]? Or did i miss something
 
  • #6
Yeah,u missed the second cosine.One from the sqrt & the other from the change of variable...

Daniel.
 
  • #7
ah that clears things up. thanks.
 

Related to Integration Help: Solving Last Problem in Multivariable Calculus

1. What is integration and why is it important in multivariable calculus?

Integration is a mathematical concept that involves finding the area under a curve. In multivariable calculus, integration is used to find the volume, surface area, and other properties of complex shapes and functions. It is an important tool for understanding and solving problems in both mathematics and other sciences.

2. How do I know when to use single or double integration in multivariable calculus?

In general, single integration is used to find the area under a curve in one variable, while double integration is used to find the volume under a surface in two variables. However, the choice between single and double integration depends on the specific problem and the variables involved. It is important to carefully read and understand the problem before deciding which type of integration to use.

3. What is the difference between definite and indefinite integration?

Definite integration involves finding the exact numerical value of the area under a curve between two specific limits. Indefinite integration, on the other hand, involves finding the general formula for the area under a curve without any specific limits. In multivariable calculus, definite integration is used to find the volume or surface area of a specific shape, while indefinite integration is used to solve more general problems.

4. How do I set up the limits of integration in multivariable calculus?

The limits of integration are determined by the boundaries of the shape or region being integrated over. This can be visualized by graphing the function or shape and identifying the minimum and maximum values for each variable. The limits of integration are typically written as a range for each variable, with the innermost variable being integrated first.

5. What are some common techniques for solving integration problems in multivariable calculus?

Some common techniques for solving integration problems include using substitution, integration by parts, and partial fractions. It is also helpful to have a good understanding of basic integration rules and properties, such as the power rule and the chain rule. Additionally, using software programs or graphing calculators can be useful for visualizing and solving complex integration problems.

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