Explanation Given Information: The provided figure is Formula used: The double integral over a nonrectangular, if R is the rectangle a ( y ) ≤ x ≤ b ( y ) and c ≤ y ≤ d , then ∫ ∫ f ( x , y ) d x d y = ∫ c d ∫ b ( y ) a ( y ) f ( x , y ) d x d y The average value of a function f ( x , y ) on the region R is f ¯ = 1 A ∬ R f ( x , y ) d x d y where A is the area of the region, A = ∬ R d x d y . The area of the triangle is 1 2 b h , where b is the base and h is the height. Calculation: Consider the figure The figure represents the region defined by four lines. Consider the line passing through ( − 2 , 0 ) and ( 0 , 2 ) . Substitute − 2 for x 1 , 0 for y 1 , 0 for x 2 and 2 for y 2 in y − y 1 y 2 − y 1 = x − x 1 x 2 − x 1 . y − 0 2 − 0 = x − ( − 2 ) 0 − ( − 2 ) y 2 = x + 2 2 y = x + 2 Thus, the equation is y = x + 2 . Consider the line passing through ( − 2 , 0 ) and ( 0 , 1 ) . Substitute − 2 for x 1 , 0 for y 1 , 0 for x 2 and 1 for y 2 in y − y 1 y 2 − y 1 = x − x 1 x 2 − x 1 . y − 0 1 − 0 = x − ( − 2 ) 0 − ( − 2 ) y = x + 2 2 y = x 2 + 1 Thus, the equation is y = x 2 + 1 . Thus, the upper limit for y is x + 2 and lower limit is x 2 + 1 and upper limit of x is 0 and lower limit is − 2 . Consider the line passing through ( 2 , 0 ) and ( 0 , 2 ) . Substitute 2 for x 1 , 0 for y 1 , 0 for x 2 and 2 for y 2 in y − y 1 y 2 − y 1 = x − x 1 x 2 − x 1 . y − 0 2 − 0 = x − 2 0 − 2 y 2 = x − 2 − 2 y = − x + 2 Thus, the equation is y = − x + 2 . Consider the line passing through ( 2 , 0 ) and ( 0 , − 1 ) . Substitute 2 for x 1 , 0 for y 1 , 0 for x 2 and − 1 for y 2 in y − y 1 y 2 − y 1 = x − x 1 x 2 − x 1 . y − 0 1 − 0 = x − 2 0 − 2 y = x − 2 − 2 y = − x 2 + 1 Thus, the equation is y = − x 2 + 1 . Thus, upper limit for y is − x + 2 and lower limit is − x 2 + 1 and upper limit for x is 2 and lower limit is 0 . Thus, the region R is − 2 ≤ x ≤ 0 and x 2 + 1 ≤ y ≤ x + 2 , 0 ≤ x ≤ 2 and − x 2 + 1 ≤ y ≤ − x + 2 . The integration is defined as, ∬ R f ( x , y ) d x d y = ∫ − 2 0 ∫ x 2 + 1 x + 2 f ( x , y ) d y d x + ∫ 0 2 ∫ − x 2 + 1 − x + 2 f ( x , y ) d y d x . Substitute x 2 for f ( x , y ) and simplify the integral ∫ − 2 0 ∫ x 2 + 1 x + 2 f ( x , y ) d y d x + ∫ 0 2 ∫ − x 2 + 1 − x + 2 f ( x , y ) d y d x with respect to y

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ISBN: 9781337514309

Author: Waner

Publisher: Cengage

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Chapter 8.5, Problem 30E

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To calculate: The average value of the function f(x,y)=x2+y2 over the region described in the figure.

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Chapter 8.5, Problem 29E

Chapter 8.5, Problem 31E

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Chapter 8 Solutions

Applied Calculus

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The average value of the function f ( x , y ) = x 2 + y 2 over the region described in the figure.

Solution Summary: The author calculates the average value of the function f(x,y)=y over the region described in the figure.

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To calculate: The average value of the function f(x,y)=x2+y2 over the region described in the figure.

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Chapter 8 Solutions