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

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MA1104 Multivariable Calculus Lecture 12 Dr KU Cheng Yeaw Thursday March 5, 2009
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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.
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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.
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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
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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 ) dx dy.
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Type I region:
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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.
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Type II region:
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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 ) dx dy.
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Polar rectangle:
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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 0 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.
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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:
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So R = { ( r, θ ) : 1 r 2 , 0 θ π } . Changing to polar coordinates for double integral, we have ZZ R (3 x + 4 y 2 ) dA = Z π 0 Z 2 1 (3 r cos θ + 4 r 2 sin 2 θ ) r dr dθ = Z π 0 Z 2 1 (3 r 2 cos θ + 4 r 3 sin 2 θ ) dr dθ = Z π 0 r 3 cos θ + r 4 sin 2 θ r =2 r =1
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ZZ R (3 x + 4 y 2 ) dA = Z π 0 (7 cos θ + 15 sin 2 θ ) = Z π 0 7 cos θ + 15 2 (1 - cos 2 θ ) = 7 sin θ + 15 θ 2 - 15 4 sin 2 θ π 0 = 15 π 2 .
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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 . So the solid lies under the paraboloid and above the circular disk D given by x 2 + y 2 1 .
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In polar coordinates, D is given by D = { ( r, θ ) : 0 r 1 , 0 θ 2 π } .
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Since 1 - x 2 - y 2 = 1 - r 2 , we have Volume = ZZ D (1 - x 2 - y 2 ) dA = Z 2 π 0 Z 1 0 (1 - r 2 ) r dr dθ = Z 2 π 0 Z 1 0 ( r - r 3 ) dr = 2 π r 2 2 - r 4 4 1 0 = π 2 .
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What happened if we had used the rectangle coordinates instead of polar coordinates? Then we would have obtained Volume = ZZ D (1 - x 2 - y 2 ) dA = Z 1 - 1 Z 1 - x 2 - 1 - x 2 (1 - x 2 - y 2 ) dy dx which is more complicated to evaluate.
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