Vector calculus - show that the integral takes the form of (0, a, 0)

[celine]

Homework Statement

suppose that V is the hemisphere with |x| <= R, z>=0, F = (z, x^2+y^2, 0) in Cartesian coordinates. Verify the equality shown below by calculating both integrals and showing that they take the form (0, a, 0) in Cartesian coordinates, for some constant a to be determined.

Relevant Equations

It was previously proven that the first equation below should hold.

Since the question asks for Cartesian coordinates, I wrote dV as 2pi(x^2+y^2+z^2)dxdydz and did the integral over the left hand side of the equation with x, y, z from 0 to R. My integral returned (0, 2*pi*R^5, 5/3*pi*R^6) which doesn't seem right.

I also tried to compute the right-hand side of the equation, letting dS = dxdy, however I'm not sure how to work with the curl between F and dS. Should I use some vector identity and rewrite this?

Thanks for your help in advance!

 

[pasmith]

"In cartesian coordinates" means to use a cartesian basis for the vectors, rather than a spherical polar basis. This is good practise, since in spherical polars the basis vectors are not constants and so you cannot integrate componentwise as you can in cartesians. The domain of integration, on the other hand, is best expressed in spherical polars.

Thus you want \int_S \mathbf{F} \times d\mathbf{S} = <br /> \int_0^{2\pi} \int_0^{\pi/2} \begin{pmatrix} z \\ x^2 + y^2 \\ 0 \end{pmatrix} \times<br /> \begin{pmatrix} x/R \\ y/R \\ z/R \end{pmatrix} R^2 \sin \theta\,d\theta\,d\phi etc.

 

[celine]

"In cartesian coordinates" means to use a cartesian basis for the vectors, rather than a spherical polar basis. This is good practise, since in spherical polars the basis vectors are not constants and so you cannot integrate componentwise as you can in cartesians. The domain of integration, on the other hand, is best expressed in spherical polars.

Thus you want \int_S \mathbf{F} \times d\mathbf{S} =<br /> \int_0^{2\pi} \int_0^{\pi/2} \begin{pmatrix} z \\ x^2 + y^2 \\ 0 \end{pmatrix} \times<br /> \begin{pmatrix} x/R \\ y/R \\ z/R \end{pmatrix} R^2 \sin \theta\,d\theta\,d\phi etc.

Thank you for the response, however I am wondering how you got the components of dS to be (x/R, y/R, z/R)? Thanks!

 

[nucl34rgg]

Keep in mind what dS represents. It should have a magnitude equal to the differential surface area and a direction pointing normal to the surface. That direction vector ought to be normalized.

 

[celine]

Keep in mind what dS represents. It should have a magnitude equal to the differential surface area and a direction pointing normal to the surface. That direction vector ought to be normalized.

Aah I see! I didn't consider the normalization! Thank you