Slide_12 - MA1104 Multivariable Calculus Lecture 12 Dr KU...

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Unformatted text preview: MA1104 Multivariable Calculus Lecture 12 Dr KU Cheng Yeaw Thursday March 5, 2009 Recap Last lecture, we learned • how to evaluate double integral over Type I and Type II region • how to evaluate double integral over polar rectangle. Overview In this lecture, we shall • evaluate double integrals over more general polar region • introduce triple integral over rectangular box, Type 1, Type 2 and Type 3 regions. Double Integrals in Polar Coordinates We have learned how to evaluate double integral over D where D is of the following type: • rectangle; • region of Type I or Type II. • polar rectangle To summarize, Fubini’s Theorem If f is continuous on the rectangle R = [ a,b ] × [ c,d ] , then ZZ R f ( x,y ) dA = Z b a Z d c f ( x,y ) dy dx = Z d c Z b a f ( x,y ) dxdy. Type I region: Integral over Type I Domain If f is continuous on a Type I domain D such that D = { ( x,y ) : a ≤ x ≤ b,g 1 ( x ) ≤ y ≤ g 2 ( x ) } then ZZ D f ( x,y ) dA = Z b a Z g 2 ( x ) g 1 ( x ) f ( x,y ) dy dx. Type II region: Integral over Type II Domain If f is continuous on a Type II domain D such that D = { ( x,y ) : c ≤ y ≤ d,h 1 ( y ) ≤ x ≤ h 2 ( y ) } then ZZ D f ( x,y ) dA = Z d c Z h 2 ( y ) h 1 ( y ) f ( x,y ) dxdy. Polar rectangle: Change to Polar Coordinates in Double Integral If f is continuous on a polar rectangle R given by R = { ( r,θ ) : 0 ≤ a ≤ r ≤ b,α ≤ θ ≤ β } where ≤ β- α ≤ 2 π , then ZZ R f ( x,y ) dA = Z β α Z b a f ( r cos θ,r sin θ ) r dr dθ. Example 1. Evaluate RR R (3 x + 4 y 2 ) dA where R is the region in the upper half-plane bounded by the circles x 2 + y 2 + 1 and x 2 + y 2 = 4 . Solution. The region R is shown below: So R = { ( r,θ ) : 1 ≤ r ≤ 2 , ≤ θ ≤ π } . Changing to polar coordinates for double integral, we have ZZ R (3 x + 4 y 2 ) dA = Z π Z 2 1 (3 r cos θ + 4 r 2 sin 2 θ ) r dr dθ = Z π Z 2 1 (3 r 2 cos θ + 4 r 3 sin 2 θ ) dr dθ = Z π r 3 cos θ + r 4 sin 2 θ r =2 r =1 dθ ZZ R (3 x + 4 y 2 ) dA = Z π (7cos θ + 15sin 2 θ ) dθ = Z π 7cos θ + 15 2 (1- cos2 θ ) dθ = 7sin θ + 15 θ 2- 15 4 sin2 θ π = 15 π 2 . Example 2. Find the volume of the solid bounded by the plane z = 0 and the paraboloid z = 1- x 2- y 2 . Solution. Notice the plane and the paraboloid intersect in the circle x 2 + y 2 = 1 ....
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This note was uploaded on 03/19/2012 for the course MATH 1104 taught by Professor Kuchengyeaw during the Spring '08 term at National University of Singapore.

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Slide_12 - MA1104 Multivariable Calculus Lecture 12 Dr KU...

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