Vector calculus double integrals

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SUMMARY

The discussion focuses on calculating the volume V of the solid under the surface defined by the equation z = 4 - x² - y², specifically over the rectangle R where 0 ≤ x ≤ 1 and 0 ≤ y ≤ 2. The integral setup provided is ∫ from 0 to 2 ∫ from 0 to 1 (4 - x² - y²) dx dy, which is confirmed as correct for this calculation. The conversation also highlights the distinction between using rectangular coordinates for this problem versus polar coordinates, which are applicable in different contexts.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Familiarity with the concept of volume under a surface
  • Knowledge of rectangular vs. polar coordinate systems
  • Basic proficiency in evaluating definite integrals
NEXT STEPS
  • Study the evaluation of double integrals using rectangular coordinates
  • Learn how to convert double integrals to polar coordinates
  • Explore applications of double integrals in calculating volumes of solids
  • Review the properties of functions defined by z = f(x, y) in three-dimensional space
USEFUL FOR

Students and educators in calculus, mathematicians focusing on multivariable calculus, and anyone interested in understanding volume calculations under surfaces in three-dimensional space.

braindead101
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Find the volume V of the solid under the surface z=4-x^2-y^2 and over the rectangle R consisting of all points (x,y) such that 0<=x<=1 and 0<=y<=2.

I have started, but am unsure if my approach is correct or not.

x = 4-x^2-y^2
[tex]\int^{2}_{0}\int^{1}_{0} 4-x^{2}-y^{2} dx dy[/tex]

is this correct?
 
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braindead101 said:
Find the volume V of the solid under the surface z=4-x^2-y^2 and over the rectangle R consisting of all points (x,y) such that 0<=x<=1 and 0<=y<=2.

I have started, but am unsure if my approach is correct or not.

x = 4-x^2-y^2
[tex]\int^{2}_{0}\int^{1}_{0} 4-x^{2}-y^{2} dx dy[/tex]

is this correct?
I could have sworn I had answered this before! I particularly remember the peculiar "x= 4- x^2- y^2" term to which I noted that it should be "z=" rather than "x= ".

But that post also asked about changing to polar coordinates! Rectangle, as here, uses "rectangular" coordinates. Disks and circles use polar coordinates.
 

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