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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.
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
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:
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:
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:
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:
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 .
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 .
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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