Se ha denunciado esta presentación.
Utilizamos tu perfil de LinkedIn y tus datos de actividad para personalizar los anuncios y mostrarte publicidad más relevante. Puedes cambiar tus preferencias de publicidad en cualquier momento.
Presentation On
“Double Integration In Polar Form
With Change In Variable”
Harsh Gupta
(130800119030)
Introductive…
• Let ƒ(r,ө) be the
function of polar
coordinates r and ө,
defined on region R
bounded by the half
lines ө=α...
• If a region S in the uv-plane is transformed
into the region R in xy-plane by the
transformation x=ƒ(u,v)=g(x,y) then th...
• ∫∫R ƒ(r,ө) dA = ∫∫R ƒ(r,ө) r dr dө
• In this type of examples x is replaced by
(r cosө) and y is replaced by (r sinө ).
And dx and dy are replaced by (r dr d...
Some Examples…
Real Life Examples..
• Velocity Profile with No Slip Boundary Condition
• Double Integrals can be used to calculate the Ma...
• Determine the volume of the region that lies
under the sphere , above the plane and inside
the cylinder .
Solution
• We ...
• The function isn’t too bad. It’s just the sphere, however, we
do need it to be in the form . We are looking at the regio...
• So, the region that we want the volume for is
really a cylinder with a cap that comes from
the sphere.
• We are definite...
Double integration in polar form with change in variable (harsh gupta)
Double integration in polar form with change in variable (harsh gupta)
Double integration in polar form with change in variable (harsh gupta)
Double integration in polar form with change in variable (harsh gupta)
Double integration in polar form with change in variable (harsh gupta)
Double integration in polar form with change in variable (harsh gupta)
Próxima SlideShare
Cargando en…5
×

Double integration in polar form with change in variable (harsh gupta)

Double Integration In Polar Form With Change In Variable

  • Inicia sesión para ver los comentarios

Double integration in polar form with change in variable (harsh gupta)

  1. 1. Presentation On “Double Integration In Polar Form With Change In Variable” Harsh Gupta (130800119030)
  2. 2. Introductive… • Let ƒ(r,ө) be the function of polar coordinates r and ө, defined on region R bounded by the half lines ө=α, ө=β and the continuous curves r=ƒ1(ө), r=ƒ2(ө), where α ≤ β and 0 ≤ ƒ1(ө) ≤ ƒ2(ө).Then the region R is as shown in the figure.
  3. 3. • If a region S in the uv-plane is transformed into the region R in xy-plane by the transformation x=ƒ(u,v)=g(x,y) then the function F(x,y) defined on R can be thought of a function F(ƒ(u,v),g(u,v)) defined on S.
  4. 4. • ∫∫R ƒ(r,ө) dA = ∫∫R ƒ(r,ө) r dr dө
  5. 5. • In this type of examples x is replaced by (r cosө) and y is replaced by (r sinө ). And dx and dy are replaced by (r dr dө). Here r comes from Jacobian solution of x= r cosө and y= r sinө . And x² + y² = r².. i.e. : x-> r cosө y-> r sinө
  6. 6. Some Examples…
  7. 7. Real Life Examples.. • Velocity Profile with No Slip Boundary Condition • Double Integrals can be used to calculate the Mass Flow of air entering the Airbox of an F1 car by using Polar Coordinates and Velocity Profiles. This is a useful real world application for double integrals, and there are many others. In the near future I hope to write a follow up with comparisons to CFD and Hagen-Poiseuille Flow. • - See more at: http://consultkeithyoung.com/content/applied- math-example-double-integral-polar-coordinates-flow-f1- airbox-inlet#sthash.OuIwgNjv.dpuf
  8. 8. • Determine the volume of the region that lies under the sphere , above the plane and inside the cylinder . Solution • We know that the formula for finding the volume of a region is, • In order to make use of this formula we’re going to need to determine the function that we should be integrating and the region D that we’re going to be integrating over.
  9. 9. • The function isn’t too bad. It’s just the sphere, however, we do need it to be in the form . We are looking at the region that lies under the sphere and above the plane • (just the xy-plane right?) and so all we need to do is solve the equation for z and when taking the square root we’ll take the positive one since we are wanting the region above the xy- plane. Here is the function. • The region D isn’t too bad in this case either. As we take points, , from the region we need to completely graph the portion of the sphere that we are working with. Since we only want the portion of the sphere that actually lies inside the cylinder given by this is also the region D. The region D is the disk in the xy-plane. • For reference purposes here is a sketch of the region that we are trying to find the volume of.
  10. 10. • So, the region that we want the volume for is really a cylinder with a cap that comes from the sphere. • We are definitely going to want to do this integral in terms of polar coordinates so here are the limits (in polar coordinates) for the region, • and we’ll need to convert the function to polar coordinates as well. • The volume is then,

×