Example// Find the volume enclosed by two surfaces Z1=2+x² +y2 ,Z2=4-x² - y² Solution // Z1 Z2 2+x² + y² = 4x² - y² 2 + x² + y² - 4 + x² + y² = 0 2x²+2y2 2 0 x² + y² = 1 ↔ x² + y² = p²² y=√1-x² Y2=-√√1-x= Volume=ff zdA=ff z dy dx = ff (Z2 - Z1) dy dx = (4 x2 y2-2-x² - y²) dy dx Volume-2 √1-2 (2- 2x² - 2y²)dy dx Or Volume-(2- 2x² – 2y²)dy dx Or Volume-41-2 ( (2-2x²-2y2)dy dx v=4 - 4 √² (2y – 2x² y — ²x²³)| √1-x² dx - = 4 + ſo (2√1 − x² - 2x² (1 − x²) - -2x²(1-x²) 2 (√1 = x²)³) dx 2(√1-x2)3 - 3
Example// Find the volume enclosed by two surfaces Z1=2+x² +y2 ,Z2=4-x² - y² Solution // Z1 Z2 2+x² + y² = 4x² - y² 2 + x² + y² - 4 + x² + y² = 0 2x²+2y2 2 0 x² + y² = 1 ↔ x² + y² = p²² y=√1-x² Y2=-√√1-x= Volume=ff zdA=ff z dy dx = ff (Z2 - Z1) dy dx = (4 x2 y2-2-x² - y²) dy dx Volume-2 √1-2 (2- 2x² - 2y²)dy dx Or Volume-(2- 2x² – 2y²)dy dx Or Volume-41-2 ( (2-2x²-2y2)dy dx v=4 - 4 √² (2y – 2x² y — ²x²³)| √1-x² dx - = 4 + ſo (2√1 − x² - 2x² (1 − x²) - -2x²(1-x²) 2 (√1 = x²)³) dx 2(√1-x2)3 - 3
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Please (re)solve the questions below in some detail.......

Transcribed Image Text:Example// Find the volume enclosed by two surfaces
Z1=2+x² +y2
,Z2=4-x² - y²
Solution //
Z1 Z2
2+x² + y² = 4x² - y²
2 + x² + y² - 4 + x² + y² = 0
2x²+2y2 2 0
x² + y² = 1 ↔ x² + y² = p²²
y=√1-x²
Y2=-√√1-x=
Volume=ff zdA=ff z dy dx = ff (Z2 - Z1) dy dx =
(4 x2 y2-2-x² - y²) dy dx
Volume-2 √1-2 (2- 2x² - 2y²)dy dx
Or Volume-(2- 2x² – 2y²)dy dx
Or Volume-41-2 (
(2-2x²-2y2)dy dx
v=4
-
4 √² (2y – 2x² y — ²x²³)| √1-x² dx
-
= 4
+ ſo (2√1 − x² - 2x² (1 − x²) -
-2x²(1-x²) 2 (√1 = x²)³) dx
2(√1-x2)3
-
3
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