15_C_1 - dS = 2 π 2 cos v d v The area of the whole given...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
1. Given: x = ( 2 + cos v) cos u , y = ( 2 + cos v) sin u , z = sin v for 0 u 2 π , 0 v π . The cylindrical coordinate r satisfies r 2 = x 2 + y 2 = ( 2 + cos v) 2 r = 2 + cos v ( r - 2 ) 2 + z 2 = 1 . This equation represents the surface of a torus, obtained by rotating about the z -axis the circle of radius 1 in the xz -plane centred at ( 2 , 0 , 0 ) . Since 0 v π implies that z 0, the given surface is only the top half of the toroidal surface. By symmetry, x = 0 and y = 0. A ring-shaped strip on the surface at angular position v with width d v has radius 2 + cos v , and so its surface area is
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: dS = 2 π( 2 + cos v) d v . The area of the whole given surface is S = Z π 2 π( 2 + cos v) d v = 4 π 2 . The strip has moment z dS = 2 π( 2 + cos v) sin v d v about z = 0, so the moment of the whole surface about z = 0 is M z = = 2 π Z π ( 2 + cos v) sin v d v = 2 π ±-2 cos v-1 4 cos ( 2 v) ²³ ³ ³ ³ π = 8 π. Thus z = 8 π 4 π 2 = 2 π . The centroid is ( , , 2 /π) ....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online