Integrate f(x,y,z)=x over the region bounded above by z=4-x^2-y^2 and below by z=x^2+3y^2

Question

Im really stuck on this one. If anyone could tell me how to do this and the answer that would be great!

 

Integrate f(x,y,z)=x over the region bounded above by z=4-x^2-y^2 and below by z=x^2+3y^2

Answer 1

 

Hi Billy,

 

This is a typical example of volume integral,

where the answer means the x-coordinate of the mass center.

 

First, we need to figure out the region

over which the function f(x, y, z) to be integrated.

 

a) Integration interval along the z-axis

 

     x² + 3y² ≤ z ≤ 4 - x² - y²   (Eq. 1, as given by the problem)

 

From Eq. 1 above, we derive the following inequality.

 

     x² + 3y² ≤ 4 - x² - y²

 

     2x² + 4y² ≤ 4

 

     (1/2)x² + y² ≤ 1                    (Eq. 2)

 

Eq. 2 represents the inside of an ellipse

over which the integrand needs to be integrated.

 

From Eq. 2, we can derive

 

     -√(1-(1/2)x²) ≤ y ≤ √(1-(1/2)x²)

 

b) Integration interval along the y-axis

 

     -√(1-(1/2)x²) ≤ y ≤ √(1-(1/2)x²)

 

c) Integration interval along the x-axis

     -√2 ≤ x ≤ √2

 

Now, we integrate f(x, y, z) over the volume

defined by a), b), and c) above.

 

     ∫^√2_-√2∫^yu_yl∫^zu_zl f(x, y, z) dzdydx

 

where zu = 4 - x² - y², zl = x² + 3y²,

          yu = √(1-(1/2)x²), yl = -√(1-(1/2)x²).

 

     ∫^√2_-√2∫^yu_yl∫^zu_zl f(x, y, z) dzdydx

 

               = ∫^√2_-√2∫^yu_yl∫^zu_zl x dzdydx

 

               = ∫^√2_-√2∫^yu_yl x (∫^zu_zl 1 dz) dydx   (Eq. 3)

 

where x is regarded as a constant when integrating over z.

 

Keep calculating Eq. 3,

 

     ∫^√2_-√2∫^yu_yl∫^zu_zl f(x, y, z) dzdydx

 

               = ∫^√2_-√2∫^yu_yl x((4-x²-y²) - (x²+3y²)) dydx

 

               = ∫^√2_-√2∫^yu_yl x(4-2x²-4y²) dydx

 

               = ∫^√2_-√2∫^yu_yl (4x-2x³-4xy²) dydx

 

               = ∫^√2_-√2 (4xy - 2x³y - (4/3)xy³)|^yu_yl dx

 

               = ∫^√2_-√2  ((4x-2x³)y - (4/3)xy³)|^yu_yl dx

 

               = ∫^√2_-√2  (4x-2x³)(yu-yl) - (4/3)x(yu³-yl³) dx

 

               = ∫^√2_-√2 (yu-yl)(4x-2x³-(4/3)x(yu²+yuyl+yl²)) dx

 

               = ∫^√2_-√2 2√(1-(1/2)x²)(4x-2x³-(4/3)x(1-(1/2)x²)) dx   (Eq. 4)

 

where yu - yl = 2√(1-(1/2)x²) and

          yu² + yuyl + yl² = 1- (1/2)x².

 

After some algebraic simplification, Eq. 4 becomes

 

     ∫^√2_-√2∫^yu_yl∫^zu_zl f(x, y, z) dzdydx

 

               = ∫^√2_-√2 (4√2/3)·x·(2-x²)^3/2 dx

 

               = -(4√2/15)(2-x²)^5/2|^√2_-√2

 

               = 0