Explanation Given: In the term ∫ 0 a ∫ 0 b e max { b 2 x 2 , a 2 y 2 } d y d x , limits a and b are positive. Formula used: Calculation: Let I = ∫ 0 a ∫ 0 b e max { b 2 x 2 , a 2 y 2 } d y d x . Divide the rectangle into two parts by the diagonal line a y = b x . On the lower triangle, b 2 x 2 ≥ a 2 y 2 because y ≤ b a x . Consider the lower triangle, Then, limits of x lies from 0 to 1 and value of y lies from 0 to b a x . Similarly, For upper triangle, The value of limits of y lies from 0 to 1 and value of x lies from 0 to a b y . The value of the integral is the sum of the integral on both the triangle. The value of the integral is: I = ∫ 0 a ∫ 0 b x a e b 2 x 2 d y d x + ∫ 0 b ∫ 0 a y b e a 2 y 2 d x d y = ∫ 0 <

Calculus, Early Transcendentals (Instructor's)

7th Edition

ISBN: 9781337552530

Author: Larson

Publisher: CENGAGE L

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Chapter 14.2, Problem 75E

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To calculate: The integral ∫0a∫0bemax{b2x2,a2y2}dydx.

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Chapter 14.2, Problem 74E

Chapter 14.2, Problem 76E

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Chapter 14 Solutions

Calculus, Early Transcendentals (Instructor's)

Ch. 14.1 - Evaluate the iterated integral: 0433cosrdrd Ch. 14.1 - CONCEPT CHECK Region of Integration Sketch the...Ch. 14.1 - Evaluate the integral: 0x(2xy)dy Ch. 14.1 - Evaluate the integral: xx2yxdy Ch. 14.1 - Evaluate the integral: 04x2x2ydy Ch. 14.1 - Evaluate the integral: x3x(x2+3y2)dy Ch. 14.1 - Evaluate the integral: eyyylnxxdx;y0 Ch. 14.1 - Evaluate the integral: 1y21y2(x2+y2)dx Ch. 14.1 - Evaluate the integral: 0x2yeyxdy Ch. 14.1 - Evaluate the integral: y2sin3xcosydx

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The integral ∫ 0 a ∫ 0 b e max { b 2 x 2 , a 2 y 2 } d y d x .

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To calculate: The integral ∫0a∫0bemax{b2x2,a2y2}dydx.

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Chapter 14 Solutions