# Thread: calculate line integral and draw curve C

1. ## calculate line integral and draw curve C

I am truly astonished to the amount of help I received, I'd like to thank Scott, Mr. Fantastic, Chris, and TheEmptySet for all this help you guys have given me. I now feel like I'm no longer alone...

This one is also a tuffy, perhaps someone may want to show me the light?

Question:

Let C be the closed curve C1 + C2 , where C1 is given by

r1 (t) = t i + t^2 j, o <= t <= 1

r2 (t) = (2 - t) i + (2 - t) j , 1 <= t <= 2

a) Draw a sketch of the curve C, with arrows to indicate orientation (im not sure if it's possible to do this online, but a description will be very very much appreciated)

b) Calculate the line integral $\displaystyle \int_C \mathbf{F}\cdot\,dr$ , where F is the vector field: F ( x, y ) = x^2 i + x y j

I am truly astonished to the amount of help I received, I'd like to thank Scott, Mr. Fantastic, Chris, and TheEmptySet for all this help you guys have given me. I now feel like I'm no longer alone...

This one is also a tuffy, perhaps someone may want to show me the light?

Question:

Let C be the closed curve C1 + C2 , where C1 is given by

r1 (t) = t i + t^2 j, o <= t <= 1

r2 (t) = (2 - t) i + (2 - t) j , 1 <= t <= 2

a) Draw a sketch of the curve C, with arrows to indicate orientation (im not sure if it's possible to do this online, but a description will be very very much appreciated)

b) Calculate the line integral $\displaystyle \int_C \mathbf{F}\cdot\,dr$ , where F is the vector field: F ( x, y ) = x^2 i + x y j
$\displaystyle C_1$ is the parabola $\displaystyle y = x^2$ starting at $\displaystyle (0,0)$ and going to $\displaystyle (1,1).$ $\displaystyle C_2$ is a straight line starting at $\displaystyle (1,1)$ and going back to $\displaystyle (0,0)$. I'll do the first and you the second.

If $\displaystyle x =t$ and $\displaystyle y = t^2$ then $\displaystyle dx = dt$ and $\displaystyle dy = 2t\,dt.$ The line integral becomes

$\displaystyle \int \limits_{c_1} x^2 dx + xy\,dy = \int_0^1 t^2\, dt + t \cdot t^2 \cdot2t\,dt = \int_0^1 r^2 + 2t^4 \, dt = \frac{11}{15}$

3. ## Thanks Danny!!!!!

Thanks Danny, ur the MANN!!! I really appreciate it However I am stuck on the second one. Since I don't know how to write in the correct notation, can you show me the second one as well?

Thanks Danny, ur the MANN!!! I really appreciate it However I am stuck on the second one. Since I don't know how to write in the correct notation, can you show me the second one as well?
Originally Posted by Danny
$\displaystyle C_1$ is the parabola $\displaystyle y = x^2$ starting at $\displaystyle (0,0)$ and going to $\displaystyle (1,1).$ $\displaystyle C_2$ is a straight line starting at $\displaystyle (1,1)$ and going back to $\displaystyle (0,0)$. I'll do the first and you the second.

If $\displaystyle x =t$ and $\displaystyle y = t^2$ then $\displaystyle dx = dt$ and $\displaystyle dy = 2t\,dt.$ The line integral becomes

$\displaystyle \int \limits_{c_1} x^2 dx + xy\,dy = \int_0^1 t^2\, dt + t \cdot t^2 \cdot2t\,dt = \int_0^1 r^2 + 2t^4 \, dt = \frac{11}{15}$
Here $\displaystyle x = 2 - t, y = 2 - t, t = 1 \to 2$ so $\displaystyle dx = dt, dy=dt$ so the line integral becomes

$\displaystyle \int_1^2 (2-t)^2 dt + (2-t)(2-t)dt$

although what I might suggest is to follow the line $\displaystyle y = x$ ($\displaystyle dy=dx$) from $\displaystyle x = 1 \to 0$ so

$\displaystyle \int_1^0 2x^2dx = - \frac{1}{3}$

5. ## I love you!!!!

Thanks soo much danny!!!