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(d) \iint_D (y^2 + 3x) dA where D is the region in the third quadrant between $x^2 + y^2 = 1$ and $x^2 + y^2 = 9$. (Hint: Convert to polar co-ordinates)

          (d) \iint_D (y^2 + 3x) dA where D is the region in the third quadrant between $x^2 + y^2 = 1$ and $x^2 + y^2 = 9$.
(Hint: Convert to polar co-ordinates)
        
(d) (y^2 + 3x) dA where D is the region in the third quadrant between x^2 + y^2 = 1 and x^2 + y^2 = 9.
(Hint: Convert to polar co-ordinates)

Added by Miguel V.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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d) y^2 + 3x dA where D is the region in the third quadrant between x + y = 1 and x^2 + y^2 = 9 (Hint: Convert to polar coordinates)
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Transcript

-
00:01 Okay, let's look at this circle where it is, okay? because it's gonna decide our boundaries here.
00:09 So, center is at negative 1 comma, 0 comma negative 1, so right here.
00:16 Ooh, look at that circle right here.
00:20 This is negative 2, this is negative 1, so what it's gonna do is it's gonna help us with the theta, okay? theta is right here.
00:28 And theta runs from pi to 2 pi.
00:33 Because remember, theta goes from here, 0 to 2 pi.
00:35 So, this is the polar coordinates, so theta is gonna run from, that's 0, pi to 2 pi.
00:48 And r is gonna run from 0 to, well, they gave you that, negative 2 sine theta...
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