Line Integrals

Tue, 29 Jul 2014 06:58:55 -0500

I'm hoping someone can show me what I am doing wrong with this. In section 16.2, question #3, it asks you to compute $\int (z\cos(xy))ds$ along the line segment from $(1,0,1)$ to $(2,2,3).$ Here is my calculations: $$P(t)=<1,0,1>+$$ $$\int_1^2 (1+2t)\cos(2t+2t^2)\sqrt{1^2+2^2+2^2}dt$$ $$u=2t+2t^2$$ $$\frac{du}{2(1+2t)}=dt$$ $$3\int_1^2 (\cos(u)(1+2t)\frac{1}{2(1+2t)})du$$ $$\frac{3}{2}\int_1^2 (\cos(u))du$$ $$\frac{3}{2}\sin(u)\biggr|_1^2$$ $$\frac{3}{2}\sin(2)-\frac{3}{2}\sin(1)$$ where did I go wrong?

Answer by Christina for Line Integrals

Tue, 29 Jul 2014 09:38:44 -0500

Anonymous, the first thing I noticed was your bounds of integration. I posted a question about this last night as I was struggling with how we know. Wes had a nice answer and Justin clarified a bit in class for me. It is all about the time interval, so typically when integrating along a line segment, you are going from t=0 to t=1, thus $\int_0^1$. Then when you use u-sub, don't forget to change your bounds. Let's start here: $$u=2t+2t^2$$ $$\frac{du}{2(1+2t)}=dt$$ When t=1, u=0. When t=1, u=4. Giving you: $$3\int_0^4 (\cos(u)(1+2t)\frac{1}{2(1+2t)})du$$ $$\frac{3}{2}\int_0^4 (\cos(u))du$$ $$\frac{3}{2}\sin(u)\biggr|_0^4$$ $$\frac{3}{2}\sin(4)-\frac{3}{2}\sin(0)$$ $$\frac{3}{2}\sin(4)-0$$ $$\frac{3\sin(4)}{2}$$ Which is the answer! Hope this helps, everything else was spot on!

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Line Integrals

Answer by Christina for Line Integrals