EXAMPLE 4 Evaluate the integral LL Solution Integration with respect to y gives VI-² which is difficult to evaluate with
Posted: Wed Jul 06, 2022 1:20 pm
- Example 4 Evaluate The Integral Ll Solution Integration With Respect To Y Gives Vi Which Is Difficult To Evaluate With 1 (85.09 KiB) Viewed 13 times
EXAMPLE 4 Evaluate the integral LL Solution Integration with respect to y gives VI-² which is difficult to evaluate without tables. Things go better if we change the original integral to polar coordinates. The region of integration in Cartesian coordinates is given by the inequalities 0 ≤ y ≤ V1 - x² and 0 ≤ x ≤ 1, which correspond to the interior of the unit quarter circle x² + y² = 1 in the first quadrant. (See Figure 15.27, first quadrant.) Substituting the polar coordinates x = r cos 0, y = r sin 0, 0≤ 0 ≤ π/2, and 0 ≤ r ≤ 1, and replacing dy dx by r dr de in the double integral, we get SS (x² + y²) dy dx. [ (2√₁ - 2 + (1 - 3²9³²) de dx, /2 √3/ (x² + y²) dy dx = (²) r dr de π/2 - /*^~-²1 - F do = The polar coordinate transformation is effective here because x² + y² simplifies to ² and the limits of integration become constants.