Question
LeAeytleaulP( Ldt C be the curve that travels on the unit circle counterclockwise from (1,0) to (0,1). Calculate Jc (Zxe" coslr') -y) dx + (c"sinfx') +x) dy: 7 Lct C bc thc cunt cas(nt )-t" 6"2" sin(7t/2)) for 0 <t<1 nd Ict F e (Jv" a Y))i coslx yli Find Fands (Hint nrall thut n Jc 4 (-# 8Lct Sbe thc Icgion + * X) pue | > .6 26 + {1 Let Ceds be the boundary of the region oriented ounter-clockwisc Find (ycosx 1 0 1 XP (6 sinx+x+e") dy LCbethe
Le Aeytleaul P( Ldt C be the curve that travels on the unit circle counterclockwise from (1,0) to (0,1). Calculate Jc (Zxe" coslr') -y) dx + (c"sinfx') +x) dy: 7 Lct C bc thc cunt cas(nt )-t" 6"2" sin(7t/2)) for 0 <t<1 nd Ict F e (Jv" a Y))i coslx yli Find Fands (Hint nrall thut n Jc 4 (-# 8Lct Sbe thc Icgion + * X) pue | > .6 26 + {1 Let Ceds be the boundary of the region oriented ounter-clockwisc Find (ycosx 1 0 1 XP (6 sinx+x+e") dy LCbethe closed cune (* - 282 + (y_ 2)2[
Answers
Evaluate the integral $\int_{C} \mathbf{F} \cdot d \mathbf{r},$ where $C$ is the boundary of the region $R$ and $C$ is oriented so that the region is on the left when the boundary is traversed in the direction of its orientation. $\mathbf{F}(x, y)=\left(e^{-x}+3 y\right) \mathbf{i}+x \mathbf{j} ; C$ is the boundary of the region $R$ inside the circle $x^{2}+y^{2}=16$ and outside the circle $x^{2}-2 x+y^{2}=3$
Problem. 13. First we drew the curve access ableto y squared And why is able through X Square as here This is X is ableto one squared and this is really different to excess square and this is a rounded abounded Eso the integration concede toe one off x squared Why bx waas y plus x y squared b Why? Using the green serum which can be written it, that every integration off Y squared minus X squared de y and V X, which is equal toe negative exposure. Seven over 21 plus export 5/5 minus 4/15 export 5/2 from 0 to 1, which is equal to negative for over 35.
Our problem. 1. 53. Using green serum. Evaluate the falling Integral. Okay, simplifying this. Get the end of a four x cubed plus four x y squared when it zero d a converting this to polar d A and then plugging in our limits. That's simplifying this integral. We got zero.
I use a Christian T Value is the line to cross eyes from radius to serve with the Radius three Circle. The first righted US area in two groups. Derivative Off this with respect go axis Reacts Square minus the ripped him off this part with respect or why is on FF three y square So we have three X Square plus y square thirty eh? And because of the natural region, it's probably easier to use to use the poorer coordinate. So accept who? Zero coz I see the why. Who's Rose scientist? I remember thie, a host roti roadies, So this integral will be Rose should be from the radio inside, which is to the radius of all side with three. And it's a four circles off the coast from zero to pie X square plus y square. I is Rose Square, the a's roti roti said. Also, we have roque you the road so to you. Sory road with a fourth over four. So this will keeps us eighty one over four minus sixteen over four. So we have ah, sixty five over four times two pi times three Teo. So we have one ninety five pie over to you