- #1

chwala

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- Homework Statement
- See attached

- Relevant Equations
- Green's theorem

My lines are as follows;

##y=\sqrt x## and ##y=x^2## intersect at ##(0,0## and ##(1,1)##.

Along ##y=\sqrt x##, from ##(0,0)## to ##(1,1)## the line integral equals,

$$\int_0^1 [3x^2-8x] dx + \dfrac{4\sqrt x-6x\sqrt x}{2\sqrt x} dx $$

$$=\int_0^1[3x^2-8x+2-3x]dx=\int_0^1[3x^2-11x+2]dx = [x^3-5.5x^2+2x]_0^1=-2.5$$

Along ##y=x^2##, from ##(1,1)## to ##(0,0)##, the line integral equals,

$$\int_1^0 [3x^2-8x^4]dx + [8x^3-12x^4]dx = \int_1^0 [8x^3-20x^4+3x^2] dx = [x^3-4x^5+2x^4]_1^0 =1$$

The required integral = ##-2.5+1=-1.5##

using, $$\int_c M dx + N dy = \iint_c \left[\dfrac{∂N}{∂x} - \dfrac{∂M}{∂y}\right] dx dy$$

...

we shall have,

$$\int_{\sqrt x}^{x^{2}} (10y)dx = [5y^2]_{\sqrt x}^{x^{2}} = 5x^4-5x $$

then

$$\int_{x=0}^1 [5x^4-5x] dx = [x^5 - 2.5x^2]_0^1 = -1.5$$

so that the theorem is verified.

Maybe my question would be; how to get the second value ##\dfrac{5}{3}##.

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