Tutorial 8 solution - NATIONAL UNIVERSITY OF SINGAPORE...

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NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA 1505 Mathematics I Tutorial 8 (Solution Notes) 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 .) Ans :4 π Preliminaries Surface Area The formula is given by a double integral (check details in the lecture notes 8.4.4): A ( S )= ±± R ² ³ ∂z ∂x ´ 2 + ³ ∂y ´ 2 +1 dA, where R is the projection of the surface onto x - y plane. Solution To apply the formula of surface area, we need to write z as a function of x and y ,and±nd out the partial derivatives. Then we need to ±nd the projection R . We know that the equation of this sphere (centered at origin (0 , 0 , 0) with radius 2) is given by ( x 0) 2 +( y 0) 2 z 0) 2 2 , i.e., x 2 + y 2 + z 2 2 , and the surface is above z =1 ,whichmeans z> 1 > 0. Thus, z = µ 2 2 x 2 y 2 = z ( x, y µ 2 2 x 2 y 2 . Then, = µ 2 2 x 2 y 2 · = 1 2 · 1 µ 2 2 x 2 y 2 · (2 2 x 2 y 2 x µ 4 x 2 y 2 = µ 2 2 x 2 y 2 · = 1 2 · 1 µ 2 2 x 2 y 2 · (2 2 x 2 y 2 y µ 4 x 2 y 2 Therefore, the surface area is given by the double integral: A ( S R - ¸ ¸ / ¹ x µ 4 x 2 y 2 º 2 + ¹ y µ 4 x 2 y 2 º 2 dA = R ² x 2 + y 2 4 x 2 y 2 dA 161
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This is a double integral, we need to transfer it to an iterated integral. Note that the variables in the integrand are of the form x 2 + y 2 , then we use polar coordinate system, i.e., rewrite the double integral as A ( S )= ± ± rdrdθ Now, let us investigate the limits of r and θ , which is from the boundary of the region R . First of all, we consider the surface: the surface is above the plane z =1 ,andispartof the sphere x 2 + y 2 + z 2 =4 . To get the boundary of the surface, we ±rst ±nd the intersection points of them. Substitute z =1into x 2 + y 2 + z 2 =4g ives x 2 + y 2 +1=4 = x 2 + y 2 =3 . which is the equation of a circle with the radius r = 3. This means the plane z = 1 intersects the sphere at a circle of radius r = 3. Notice that the region R is the projection of the surface, which is a spherical cap, so the region is a disk of radius r = 3. Thus, in polar coordinates, the region R is given by 0 r 3 , 0 θ 2 π. ± ² ³ ´ µ 162
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Therefore, A ( S )= ±± R ² x 2 + y 2 4 x 2 y 2 +1 dA = ± 2 π 0 ± 3 0 ³ r 2 4 r 2 rdrdθ (Polar Coordinates) = ± 2 π 0 ± 3 0 ³ r 2 +(4 r 2 ) 4 r 2 = ± 2 π 0 ± 3 0 ³ 4 4 r 2 = ± 2 π 0 ± 3 0 2 r 4 r 2 drdθ = ± 2 π 0 ± 3 0 1 4 r 2 dr 2 ( dr 2 =2 rdr ) = ± 2 π 0 ± 3 0 (4 r 2 ) 1 2 dr 2 = ± 2 π 0 ± 3 0 (4 r 2 ) 1 2 d (4 r 2 ) ´ dr 2 = d ( r 2 d (4 r 2 ) µ = ± 2 π 0 2 · 4 r 2 ¸ r = 3 r =0 = ± 2 π 0 2 ( 4 3 4 0 ) ¸ = ± 2 π 0 2 =4 π.
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This note was uploaded on 05/05/2009 for the course MA MA1505 taught by Professor Ma1505 during the Spring '09 term at National University of Juridical Sciences.

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Tutorial 8 solution - NATIONAL UNIVERSITY OF SINGAPORE...

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