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NewtonApple
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Homework Statement
Evaluate the line integral [itex]∫^{(1,0)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex] along
(a) a straight line from (0,1) to (1,2);
(b) the parabola x=t, y=t2 + 1;
(c) straight lines from (0,1) to (1,1) and then from (1,1) to (1,2).
Homework Equations
Equation of line: y = mx + c
The Attempt at a Solution
(a) [/B]To convert given integral into one variable I used Equation of line
y=mx + c
where slope m = [itex]\frac {y_2 - y_1}{x_2 - x_1} = \frac{2-1}{1-0}=1[/itex]
and y intercept c = 1
which gives us y = x+1
dy = dx
Thus given integral
[itex]∫^{(1,0)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex]
becomes one variable x dependent
[itex] = ∫^{1}_{0} (x^2-x-1)dx + (({x+1})^2+x)dx[/itex]
[itex] = ∫^{1}_{0} (x^2-x-1)dx + (x^2+2x+1+x)dx[/itex]
[itex] = ∫^{1}_{0} (x^2-x-1)dx + (x^2+3x+1)dx[/itex]
[itex] = ∫^{1}_{0} (x^2-x-1+x^2+3x+1)dx[/itex]
[itex] = ∫^{1}_{0} (2x^2+2x)dx[/itex] ------------------ (1)
solving
[itex] \frac{2x^3}{3} +\frac{2x^2}{2}[/itex]
applying limits gives
[itex]\frac{2}{3} + 1 = \frac{2+3}{3} = \frac{5}{3}[/itex]
(b) need some hints for this part
(c) straight lines from (0,1) to (1,1) and then from (1,1) to (1,2).
from equation (1) above
[itex] = ∫^{1}_{0} (2x^2+2x)dx[/itex]
solving
[itex] \frac{2x^3}{3} +\frac{2x^2}{2}[/itex]
from (0,1) to (1,1)
we use only x: 0 to 1
we get [itex]\frac{2}{3} + 1 = \frac{2+3}{3} = \frac{5}{3}[/itex]
from (1,1) to (1,2)
we use only x: 1 to 1
we get [itex]\frac{2}{3} - \frac{2}{3} + 1 - 1 = 0 [/itex]
hence adding both parts gives [itex] \frac{5}{3} [/itex]Please tell me am I on the right track?
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