How can I solve a volume integral question with a trig substitution?

In summary, the conversation centered around a volume integral question and the attempted solutions using integration by substitution and integration by parts. The integral in question was ∫^{1}_{-1} 4(√(1-x^{2}))(x+1)dx, and the final answer was 2pi. The participants suggested using a trig substitution, specifically x=sinu, to solve the integral. The conversation ended with the confirmation that the suggested substitution led to the correct solution.
  • #1
ppy
64
0
Hi,

I was attempting a volume integral question out of a book. I know what the final answer is and what integral i am supposed to work out but I do not know how I am supposed to solve it. I have tried different ways such as integration by substitution and integration by parts but I do not seem to be getting anywhere.

This is the ∫[itex]^{1}_{-1}[/itex] 4(√(1-x[itex]^{2}[/itex]))(x+1)dx The answer is 2pi.

any help would be great thanks.
 
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  • #2
This is not a volume integral. Put x=sin u and proceed
 
  • #3
The integral started out as a triple volume integral. The part mentioned is the final part of the triple integral. I have tried x=sinu and still got nowhere. Is it a simple substitution or do I have to integrate by parts as well ?
Thanks
 
  • #4
hi ppy! :smile:
ppy said:
I have tried x=sinu and still got nowhere. Is it a simple substitution or do I have to integrate by parts as well ?

davidmoore63@y's :smile: substitution does work …

show us how far you got with it :wink:
 
  • #5
Thanks. I've got there now !
 
  • #6
The presence of the ##\sqrt{1 - x^2}## factor suggests that a trig substitution is called for, and that's the direction that davidmoore and tiny-tim are recommending.

BTW, there are no such words in English as "intergral", "intergrate", or "intergration".
 

What is a volume integral?

A volume integral, also known as a triple integral, is a mathematical concept used in calculus to find the volume of a three-dimensional shape by integrating over a given region.

How is a volume integral different from a regular integral?

A regular integral is used to find the area under a curve in two dimensions, while a volume integral is used to find the volume of a three-dimensional shape.

When is a volume integral used in real life?

Volume integrals are used in various fields such as physics, engineering, and economics to calculate quantities such as mass, pressure, and profit, respectively.

What are the steps to solve a volume integral?

To solve a volume integral, you first need to define the limits of integration for each variable, then set up the integral using the appropriate formula, and finally evaluate the integral using techniques such as substitution or integration by parts.

What are some common applications of volume integrals?

Some common applications of volume integrals include finding the volume of a solid object, calculating the total mass of a three-dimensional object, and determining the center of mass of a given shape.

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