Ngô Quốc Anh

December 6, 2007

Surface integral: The symmetric property

Filed under: Các Bài Tập Nhỏ, Giải Tích 5 — Ngô Quốc Anh @ 4:59

Why is \iint_S(x^2+y^2-2z^2)dA=0 for S being the unit sphere.

\displaystyle \int_S x^2\,dA=\int_S y^2\,dA=\int_S z^2\,dA=C,

by rotational symmetry of the sphere. The integral is

\displaystyle C+C-2C=0.

And the easiest way to find C (should you want to) is again by symmetry

\displaystyle 3C=\iint_S(x^2+y^2+z^2)\,dS=\iint_S 1\,dS=4\pi,

hence C=\frac{4\pi}{3}.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at WordPress.com.

%d bloggers like this: