enter image description here" />enter image description here" />
Evaluate the line integral by two methods: (a) directly and (b) using Green’s Theorem So I thought I knew how to do this problem but when I did it directly, the areas I got for each line were 0+2/3+4, but the overal area in the answer key is 2/3. I double checked the entire process twice when I got the 4. Maybe I'm parameterizing it wrong or used the wrong boundaries, but I was sure it was supposed to be from 0 to 1 when you parameterize in terms of t. my work: r(t1) = + t = x=t, y=0 dx=1dt, dy=0 so integral C = 0 r(t2) = + t = x=1, y=2t dx=0, dy=2dt so integral C2 = integral C(0+8t^3*2)dt from 0 to 1 = 4 r3(t) = + t = x=1-t, y=2-2t dx = -1dt, dy = -2dt integral c3 = integral c3( (1-t)(2-2t)(-1)dt + (1-t)^2(2-2t)^3(-2)dt the left integral of this evaluated to 2/3 and the right I used an algebra calculator to simplify and it evaluated to 0. So the whole thing came out to 2/3 + 4, but its really 2/3. Also, when I used green's theorem I ended up getting 6, but again its still supposed to be 2/3. I used the double integral of (dQ/dx -dP/dy)dydx with y going from 0 to 2x ( i got 2x with y=mx+b) and x going from 0 to 1. and I get 6. Am I doing something fundamentally wrong? Can someone show me every detail so I can see where I went wrong? Thanks. my work: double integral(dQ/dx -dP/dy)dydx = integral(0 to 1)integral(0 to 2x)(2xy^3 - x)dxdy = 2xy^4/4 -xy ]2x to 0 2x(2x)^4 - 2x^2 8x^5 -2x^2 ] 1 to 0 = 6
2316354654
asked Dec 4, 2017 at 3:18
2316354654 2316354654
529 3 3 gold badges 22 22 silver badges 43 43 bronze badges
$\begingroup$ Can you please post your work. so we can see what you did. Thanks. $\endgroup$
Commented Dec 4, 2017 at 3:19
$\begingroup$ i put it up, itll take me a while to find the latex commands and apply them all $\endgroup$
Commented Dec 4, 2017 at 4:02
