Explanation Given: The double integral ∫ 0 2 ∫ y 8 − y 2 1 1 + x 2 + y 2 d x d y . Concept Used: The conversion from cartesian coordinates to polar coordinates is given by: x = r cos θ , y = r sin θ , x 2 + y 2 = r 2 , d y d x = r d r d θ Calculation: From the limit of integrations, we can see that the region of integration is bounded by the inequalities 0 ≤ y ≤ 2 , and y ≤ x ≤ 8 − y 2 . The region bounded by these inequalities shown below: Observe that the region of integration is the part of circle x 2 + y 2 = 8 bounded between x -axis and the line y = x . The circle, x 2 + y 2 = 8 in polar coordinates can be written as, x = 8 − y 2 ⇒ x 2 + y 2 = 8 ⇒ r 2 = 8 ⇒ r = 2 2 And the line in polar coordinates gives, y = x r sin θ = r cos θ tan θ = 1 θ = π 4 So, the limit of integration in polar coordinates would be 0 ≤ r ≤ 2 2 , and 0 ≤ θ ≤ π 4 So, the given double integral in polar coordinates becomes ∫ 0 2 ∫ y 8 − y 2 1 1 + x 2 + y 2 d x d y = ∫ 0 π / 4 ∫ 0 2 2 1 1 + r 2 ⋅ r d r d θ Now to evaluate this integral, substitute 1 + r 2 = u ⇒ 2 r d r = d u ⇒ r d r = 1 2 d u So, the limits of integration will change accordingly as at r = 0 , <

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ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 12.3, Problem 38PS

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Evaluate the integral ∫02 ∫ y 8− y 2 1 1+ x 2 + y 2 dx dy , by converting to polar coordinates.

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Chapter 12.3, Problem 37PS

Chapter 12.3, Problem 39PS

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Evaluate the integral ∫ 0 2 ∫ y 8 − y 2 1 1 + x 2 + y 2 d x d y , by converting to polar coordinates.

Solution Summary: The author evaluates the integral by converting it to polar coordinates. The region of integration is bounded by the inequalities.

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Evaluate the integral ∫02 ∫ y 8− y 2 1 1+ x 2 + y 2 dx dy , by converting to polar coordinates.

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Chapter 12 Solutions