Mark's Calc III - Individual question feed

Section 16.3

Anonymous — Wed, 30 Jul 2014 07:15:50 -0500

Number 8 in section 16.3 asks, Evaluate $\int (10x^4=2xy^3)dx-3x^2y^2dy$ where $C$ is the part of the curve $x^5-5x^2y^2-7x^2=0$ from $(0,0)$ to $(3,2)$. Can someone explain how to solve something like this?

Section 16.2

Anonymous — Tue, 29 Jul 2014 10:06:31 -0500

I'm having trouble remembering how to parametrize a line when given 3 points. In section 16.2, #10, it asks to compute $\int <1/xy,1/(x+y)>\cdot dr$ along the path from (1,1) to (3,1) to (3,6) using straight line segments. I'm really lost on what to do here.

Line Integrals

Anonymous — 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?

How do I get the bounds of integration for line integrals?

Christina — Mon, 28 Jul 2014 12:06:32 -0500

I am currently working on the homework for 16.2. Number one, for example, states: **"Compute $\int\limits_Cxy^2ds$ along the line segment from (1,2,0) to (2,1,3)"**. I am setting it up as follows: For my parameterized line I get: $$\vec{p}(t) = \langle1,2,0\rangle + t\langle1,-1,3\rangle$$ Giving me: $$x=t+1$$ $$ x' = 1$$ $$y=2-t$$ $$ y'= -1$$ $$z=3t$$ $$ z'= 3$$ Then setting up an integral to compute: $$\int(t+1)(2-t)^2\sqrt{(1)^2+(-1)^2+(3)^2}dt$$ What I am not sure of is how to know what my bounds of integration are. Do I plug my point values in for x, y, and z? Or in this case, for x and y and then solve for t? Or just for x? I would love any help on this or to know if what I have done so far is correct or not correct...Thanks!