I previously asked a question on line integrals. That one I was just understanding how to decipher the syntax of the question itself. Now I have a new question
$$\int^{(8,3,2)}_{(0,0,0)}(2xy^2-2xz^2)dx+2x^2ydy-2x^2zdz$$
I tried to parameterize it as such
$x=t^2, dx=2tdt$
$y=t^3, dy=3t^2dt$
$z=t^5, dz=5t^4dt$
I then went ahead to try to solve this as such:
$$4\int^8_0{t^9dt}+6\int^3_0{t^9dt}-10\int^2_0{t^{13}dt}$$
Needless to say, my answer was wrong. Way wrong. The correct answer is 320 but I'm not sure how to even start heading in that direction.
Like any good juvenile delinquent you should blame society for your mistakes, and here you would be quite justified in doing so. The notation they use in calculus textbooks for line integrals is ugly and obscures what you are supposed to evaluate, which is the integral of the vector field $$\vec F=\langle2xy^2-2xz^2,2x^2y,-2x^2z\rangle$$ along an unspecified path from $\langle0,0,0\rangle$ to $\langle8,3,2\rangle$. Although the path is unspecified it has to start and end at the right places and the single parameter that defines $x$, $y$, and $z$ takes on only one value at the start of the path and one value at the end of the path. Here you have $0\le t\le3$ for what looks like a $y$-subintegral which would have $y$ ranging from $0$ to $27$, clearly wrong and the $x$-subintegral is even wronger.
First you want to decide on a path; let's choose a straight line from $\langle0,0,0\rangle$ to $\langle8,3,2\rangle$. Then along the path $$\vec r=\langle0,0,0\rangle+t\left(\langle8,3,2\rangle-\langle0,0,0\rangle\right)=\langle8t,3t,2t\rangle$$ As $t$ goes from $0$ to $1$. Check now that $\vec r(0)=\langle0,0,0\rangle$ and $\vec r(1)=\langle8,3,2\rangle$. Then $$d\vec r=d\langle8t,3t,2t\rangle=\langle8,3,2\rangle dt$$ and $$\vec F=\langle2xy^2-2xz^2,2x^2y,-2x^2z\rangle=\langle80t^3,384t^3,-256t^3\rangle$$ Along the path, so $$\int_C\vec F\cdot d\vec r=\int_0^1\langle80t^3,384t^3,-256t^3\rangle\cdot\langle8,3,2\rangle dt=\int_0^11280t^3dt=320$$ Now, the fact that no path was specified should have been a hint that the integral was independent of path so that $\vec F=\vec{\nabla}\phi$ for some potential function $\phi(x,y,z)$. So $$\frac{\partial\phi}{\partial x}=2xy^2-2xz^2$$ So $\phi=x^2y^2-x^2z^2+f(y,z)$. Then $$\frac{\partial\phi}{\partial y}=2x^2y+\frac{\partial f}{\partial y}=2x^2y$$ So $f(y,z)=g(z)$ and $\phi=x^2y^2-x^2z^2+g(z)$. Then $$\frac{\partial\phi}{\partial z}=-2x^2z+\frac{dg}{dz}=-2x^2z$$ So $g(z)=K$, a constant, and $\phi(x,y,z)=x^2y^2-x^2z^2+K$. Then $$\int_C\vec F\cdot d\vec r=\int_C\vec{\nabla}\phi\cdot d\vec r=\left.\phi(x,y,z)\right|_{\langle0,0,0\rangle}^{\langle8,3,2\rangle}=\phi(8,3,2)-\phi(0,0,0)=320+K-0-K=320$$