Setting up Double Intergrals to Calculate Volume

  • Thread starter GogumaDork
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  • #1

Homework Statement


Set up the double integral in rectangular coordinates for calculating the volume.
The solid is created by the equation
pPQYP.png



Homework Equations


Fubini's Thereom ∫R∫f(x,y)dA = ∫ab∫f(x,y)dydx


The Attempt at a Solution


The shape created by Z = 30 - x2 - y2 is 3d parabola facing downards.
Since it is bounded by Z = 5, the base shape is a circle at origin with radius 5 since
5 = 30 - x2 - y2 is equal to x2 + y2 = 25
As a result x is bounded by -5 and 5 while y is bounded by -√(25-x^2) and √(25-x^2)

As a result to set up the double integral
-55√(25-x^2)√(25-x^2)30-x^2-y^2dydx

Are the integral boundaries in correct?
 
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Answers and Replies

  • #2
uart
Science Advisor
2,797
21

Homework Statement


Set up the double integral in rectangular coordinates for calculating the volume.
The solid is created by the equation
pPQYP.png



Homework Equations


Fubini's Thereom ∫R∫f(x,y)dA = ∫ab∫f(x,y)dydx


The Attempt at a Solution


The shape created by Z = 30 - x2 - y2 is 3d parabola facing downards.
Since it is bounded by Z = 5, the base shape is a circle at origin with radius 5 since
5 = 30 - x2 - y2 is equal to x2 + y2 = 25
As a result x is bounded by -5 and 5 while y is bounded by -√(25-x^2) and √(25-x^2)

As a result to set up the double integral
-55-√(25-x^2)√(25-x^2)30-x^2-y^2dydx

Are the integral boundaries in correct?

Yes the boundaries are correct (except for the typo where you left out the "-" in the lower limit).

However, you really should be making use of the cylindrical symmetry here and changing to cylindrical coordinates. It makes the problem much easier.
 
Last edited:
  • #3
Thanks for the response! However, that is the answer I entered into Webwork but it states it is incorrect.
 
Last edited:
  • #4
Okay I managed to figure it out after reading: http://mathinsight.org/double_integral_volume

My boundaries were correct but the equation I was integrating was not.
It explained that the integral equation is not f(x,y) but f(x,y) - g(x,y), in my case g(x,y) being the "other surface" or z = 5

As a result I subtracted 5 from f(x,y) which was 30 - x^2 - y^2 and resulted in a final answer of
-55-√(25-x^2)√(25-x^2) 25-x^2-y^2dydx
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,847
967
Yes, what you had was the volume between the surface [itex]z= 30- x^2- y^2[/itex] and the xy-plane, z= 0. Unfortunately, so many basic problem ask for just the "area below the curve" and "the volume below the surface", we may develop the very bad habit of automatically assuming a lower bound of 0.

By the way, in cylindrical coordinates, that uart referred to, the upper surface is given by [itex]z= 30- r^2[/tex], the lower surface by z= 5, and the bounding circe by r= 5. The integral would be
[tex]\int_{\theta= 0}^{2\pi}\int_{r= 0}^5 (30- r^2- 5) rdrd\theta=\int_{\theta= 0}^{2\pi}\int_{r= 0}^5 (25- r^2) rdrd\theta[/tex]
 

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