dstefan
12-16-2007, 01:28 PM
Chapter 2, Problem 1:
Compute the integral of the function f(x,y) = x^2 - 2y on the region bounded by the parabola y = (x+1)^2 and the line y = 5x-1.
Solution:
Find the points where the curves y = (x+1)^2 and y = 5x-1 intersect, i.e., solve (x+1)^2 = 5x-1. The only solutions are x=1 and x =2.
The region where we integrate f(x,y) is 1 \le x \le 2, (x+1)^2 \le y \le 5x-1. Therefore,
\int_1^2 ( \int_{(x+1)^2}^{5x-1} (x^2 - 2y) dy ) dx = \int_1^2 \left( \left. (x^2y - y^2)\right|_{(x+1)^2}^{5x-1} \right) dx = \int_1^2 x^2 ( (5x-1) - (x+1)^2 ) - ( (5x-1)^2 - (x+1)^4 ) dx
= \int_1^2 ( (5x-1) - (x+1)^2 ) ( x^2 - ( (5x-1) + (x+1)^2 ) ) dx = \int_1^2 ( -x^2+3x-2 ) (-7x) dx = - \frac{7}{4}
Compute the integral of the function f(x,y) = x^2 - 2y on the region bounded by the parabola y = (x+1)^2 and the line y = 5x-1.
Solution:
Find the points where the curves y = (x+1)^2 and y = 5x-1 intersect, i.e., solve (x+1)^2 = 5x-1. The only solutions are x=1 and x =2.
The region where we integrate f(x,y) is 1 \le x \le 2, (x+1)^2 \le y \le 5x-1. Therefore,
\int_1^2 ( \int_{(x+1)^2}^{5x-1} (x^2 - 2y) dy ) dx = \int_1^2 \left( \left. (x^2y - y^2)\right|_{(x+1)^2}^{5x-1} \right) dx = \int_1^2 x^2 ( (5x-1) - (x+1)^2 ) - ( (5x-1)^2 - (x+1)^4 ) dx
= \int_1^2 ( (5x-1) - (x+1)^2 ) ( x^2 - ( (5x-1) + (x+1)^2 ) ) dx = \int_1^2 ( -x^2+3x-2 ) (-7x) dx = - \frac{7}{4}