# Tutorial8 - MA1505 Tutorial VIII Hou Likun...

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MA 1505 Tutorial VIII Hou Likun g0800870@nus.edu.sg Department of Mathematics National University of Singapore Hou Likung0800870@nus.edu.sg MA 1505 Tutorial VIII

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Question 1. Find the area of the surface consisting of the part of the sphere of radius 2 centered at origin that lies above the horizontal plane z = 1 . (Equation of this sphere is given by x 2 + y 2 + z 2 = 2 2 .) Hou Likung0800870@nus.edu.sg MA 1505 Tutorial VIII
Question 1. Find the area of the surface consisting of the part of the sphere of radius 2 centered at origin that lies above the horizontal plane z = 1 . (Equation of this sphere is given by x 2 + y 2 + z 2 = 2 2 .) Formula : For an explicit surface given by z = f ( x,y ), where ( ) lies in some region R of the x - y plane (i.e. ( ) R ), then the area of this surface is given by the following formula Area = ZZ R q 1 + f 2 x + f 2 y dxdy. Hou Likung0800870@nus.edu.sg MA 1505 Tutorial VIII

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Sol . Let z = p 2 2 - x 2 - y 2 . Then z x = - x (4 - x 2 - y 2 ) - 1 / 2 and z y = - y (4 - x 2 - y 2 ) - 1 / 2 . Substitute z = 1 into x 2 + y 2 + z 2 = 4 gives x 2 + y 2 + 1 = 4 x 2 + y 2 = 3 which is the equation of a circle of radius 3. This means the plane z = 1 intersects the sphere at a circle of radius 3. Hence the projected region R of the part of the sphere is a disk of radius 3. In polar coordinates, this is given by 0 r 3 , 0 θ 2 π. Hou Likung0800870@nus.edu.sg MA 1505 Tutorial VIII
Thus, A ( S ) = ZZ R s x 2 + y 2 4 - x 2 - y 2 + 1 dA = Z 2 π 0 Z 3 0 ± r 2 4 - r 2 + 1 ² 1 2 r drdθ = Z 2 π 0 Z 3 0 2 r (4 - r 2 ) - 1 2 drdθ = Z 2 π 0 h - 2(4

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## This note was uploaded on 05/10/2011 for the course MATH 1505 taught by Professor Yap during the Winter '11 term at National University of Singapore.

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Tutorial8 - MA1505 Tutorial VIII Hou Likun...

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