# 11.28 - Section 13.6 wrap-up Stewart computes the surface...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Section 13.6 wrap-up: Stewart computes the surface area of a sphere of radius a using a spherical coordinate parametrization. We’re doing it a different way, with polar coordinates instead. Last time we parametrized the top half of the sphere of radius a by r = 〈 r cos θ , r sin θ , sqrt( a 2 – r 2 ) 〉 with ( r , θ ) varying over D = [0, a ] × [0,2 π ]. We computed that | r r × r θ | = ar / sqrt( a 2 – r 2 ), so (picking up from where we left off) we get A ( S ) = ∫ ∫ D | r r × r θ | dA = … ..?.. ∫ 2 π ∫ a ( ar / sqrt( a 2 – r 2 )) r dr d θ ? (take a vote) No! In our formula for surface area, dA refers to area in the u , v plane (which here is the r , θ plane), NOT the x , y plane! So the area element is just dr d θ . A ( S ) = ∫ 2 π ∫ a ( ar / sqrt( a 2 – r 2 )) dr d θ = ( ∫ 2 π d θ ) ( ∫ a ( ar / sqrt( a 2 – r 2 )) dr ) = (2 π ) (– a sqrt( a 2 – r 2 ) | a ) = (2 π ) ( a 2 ) which agrees with the fact that the surface area of the sphere of radius a is 4 π a 2 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

11.28 - Section 13.6 wrap-up Stewart computes the surface...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online