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10 Parametric Surfaces and Surface Integrals Free Online FlashCards

Study 10 Parametric Surfaces and Surface Integrals with 12 free online flashcards. Review key terms, definitions, and concepts with this interactive flashcard deck.

12 cards
01
Front

Why do parametrized surfaces generally use two parameters?

Back

A surface is two-dimensional, so its points generally require two parameters, such as uu and vv, in a vector-valued parametrization r(u,v)\mathbf r(u,v).

02
Front

What is the parameter domain?

Back

The parameter domain is the region DD in the uvuv-plane over which the parameters vary to generate the surface.

03
Front

How are coordinate curves obtained from r(u,v)\mathbf r(u,v)?

Back

Holding uu constant and varying vv gives one family of coordinate curves; holding vv constant and varying uu gives the other.

04
Front

What condition makes a parametrization regular?

Back

The parametrization is regular at a point when ru×rv≠0\mathbf r_u\times\mathbf r_v\ne\mathbf 0. Then the tangent directions are not parallel and determine a tangent plane.

05
Front

How is the tangent plane found from a parametrization?

Back

At P=r(u0,v0)P=\mathbf r(u_0,v_0), a tangent-plane equation is (ru×rv)∣(u0,v0)⋅⟨x−x0,y−y0,z−z0⟩=0(\mathbf r_u\times\mathbf r_v)|_{(u_0,v_0)}\cdot\langle x-x_0,y-y_0,z-z_0\rangle=0.

06
Front

What are the two unit normals to an oriented parametrized surface?

Back

The two unit normals are N=±ru×rv∥ru×rv∥\mathbf N=\pm\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}. Choosing the sign fixes the surface orientation.

07
Front

How is the graph z=f(x,y)z=f(x,y) parametrized?

Back

A graph z=f(x,y)z=f(x,y) can be parametrized by r(x,y)=⟨x,y,f(x,y)⟩\mathbf r(x,y)=\langle x,y,f(x,y)\rangle over its region DD.

08
Front

What is the surface-area element for a parametrized surface?

Back

For r(u,v)\mathbf r(u,v), the surface-area element is dS=∥ru×rv∥ du dvdS=\|\mathbf r_u\times\mathbf r_v\|\,du\,dv.

09
Front

What is the area formula for a graph?

Back

For z=f(x,y)z=f(x,y), dS=1+fx2+fy2 dAdS=\sqrt{1+f_x^2+f_y^2}\,dA, so Area⁡(S)=∬D1+fx2+fy2 dA\operatorname{Area}(S)=\iint_D\sqrt{1+f_x^2+f_y^2}\,dA.

10
Front

How is a scalar surface integral computed parametrically?

Back

A scalar surface integral is ∬Sg dS=∬Dg(r(u,v))∥ru×rv∥ du dv\iint_S g\,dS=\iint_D g(\mathbf r(u,v))\|\mathbf r_u\times\mathbf r_v\|\,du\,dv.

11
Front

What physical quantities can a scalar surface integral represent?

Back

If gg is surface density, then ∬Sg dS\iint_S g\,dS gives total mass. If g=1g=1, it gives the surface area.

12
Front

What is the parametrized formula for flux?

Back

The flux is ∬SF⋅N dS=∬DF(r(u,v))⋅(ru×rv) du dv\iint_S\mathbf F\cdot\mathbf N\,dS=\iint_D\mathbf F(\mathbf r(u,v))\cdot(\mathbf r_u\times\mathbf r_v)\,du\,dv.