Help me fast......I will give Upvote..... Evaluate the integral by making an appropriate change of variablesY dA, where R is the region enclosed by x+y=0, X+y=4, x-y%31,and x-y36

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Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Help me fast......I will give Upvote.....

Evaluate the integral by making an appropriate change of variables
Y dA, where R is the region enclosed by x+y=0, X+y=4, x-y%31,
and x-y36

Expert Solution

Step 1

$G i v e n \iint e^{x^{2} - y^{2}} d A w h e r e R i s g i v e n x + y = 0, x + y = 4, x - y = 1, x - y = 6 l e t u = x + y a n d v = x - y t h e r e f o r e u = 0 t o u = 4 a n d v = 1 t o v = 6 |J| = |\begin{matrix} u_{x} & u_{y} \\ v_{x} & v_{y} \end{matrix}| = |\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix}| = - 1 - 1 = - 2 i n t e g r a t i o n o f c h a n g e o f v a r i a b l e i s \iint f (x, y) d A = \iint f (u, v) |J| d u d v \iint e^{x^{2} - y^{2}} d A = \int_{1}^{6} \int_{0}^{4} e^{u v} (- 2) d u d v = - 2 \int_{1}^{6} {[\frac{e^{u v}}{v}]}_{0}^{4} d v = - 2 \int_{1}^{6} (\frac{e^{4 v}}{v} - \frac{1}{v}) d v = - 2 [\int_{1}^{6} \frac{e^{4 v}}{v} d v - \int_{1}^{6} \frac{1}{v} d v] p u t 4 v = z \Rightarrow d v = \frac{1}{4} d z v = 1 \Rightarrow z = 4 a n d v = 6 \Rightarrow z = 24 = - 2 [\int_{4}^{24} \frac{e^{z}}{z} d z - \log 6 + \log 1] = - 2 [\int_{4}^{24} \frac{1 + z + \frac{z^{2}}{2!} + \frac{z^{3}}{3!} + . . .}{z} d z - \log 6] = - 2 [\int_{4}^{24} \frac{1}{z} + 1 + \frac{z}{2!} + \frac{z^{2}}{3!} + \frac{z^{3}}{4!} + . . . - \log 6] = - 2 [{\{\log z + z + \frac{z^{2}}{2.2!} + \frac{z^{3}}{3.3!} + \frac{z^{4}}{4.4!} + . . .\}}^{24}_{4} - \log 6] = - 2 [\log 24 + 24 + \frac{24^{2}}{2.2!} + \frac{24^{3}}{3.3!} + . . . - \log 4 - 4 + \frac{4^{2}}{2.2!} + \frac{4^{3}}{3.3!} + . . . - \log 6] = - 2 [4 (6 - 1) + \frac{4^{2} (6^{2} - 1)}{2.2!} + \frac{4^{3} (6^{3} - 1)}{3.3!} + . . .]$

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ISBN:

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