I have two questions about line integrals.

Find the integral below along the line of intersection of the two planes $x-y+z=0$ and $x+y+2z=0$ from the origin to the point $(3,1,-2)$ $$\int_{\mathcal C} x^2ds $$

Assistant has said "you should give priority to point if question has and parametrize wrt the point" and she has done this :

$r(t) = 3t\hat i+t\hat j -2t\hat k$ , $( 0 \le t \le1)$ (I didn't understand why t is between 0 and 1 here)

Find the integral below along the first octant part $\mathcal C$ of the curve of intersection of $z=2-x^2-2y^2$ and $z=x^2$ between $(0,1,0)$ and $(1,0,1)$ $$\int_{\mathcal C} xyds $$

$x=t,y=\sqrt{1-t^2},z=t^2 $ , $( 0 \le t \le1)$ (I didn't understand why t is between 0 and 1 again)

For 2nd question she has parametrized wrt equations although question has point. Why we did not use points for parametrization? Could someone explain difference in two questions please in the easiest way? Thanks a lot

  • $\begingroup$ I've tried to clarify your doubts, even if I'm not completely sure what exactely is not clear to you. $\endgroup$
    – user
    Dec 31 '17 at 13:16
  • $\begingroup$ @gimusi parametrization is not clear for me. For first question she multiplied point with t directly but for second one she parametrized wrt curves. I parametrized 1st one wrt curves I have obtain same parametrization and I couldn’t be sure it is true or only a coincidence? $\endgroup$
    – usereb
    Dec 31 '17 at 13:20
  • $\begingroup$ I understand your doubts. Anyway for the line segment the parametrization is very easy to find. For the curve it is more difficult and you need to pay more attention. $\endgroup$
    – user
    Dec 31 '17 at 13:37
  • $\begingroup$ @gimusi thanks for answer $\endgroup$
    – usereb
    Jan 2 '18 at 10:41
  • $\begingroup$ You are welcome, if you are ok you can accept the answer and set it as solved, bye! math.meta.stackexchange.com/questions/3286/… $\endgroup$
    – user
    Jan 2 '18 at 14:07

For the first:

The intersection of the two planes is a line segment from the origin to the point (3,1,-2). The parametric equation of the line segment is indeed:

$$r(t) = 3t\hat i+t\hat j -2t\hat k$$

  • t=0 corresponds to the origin
  • t=1 correspond to the point (3,1,-2)

In general given two points P and Q the parametrization for a line segment PQ is obtained by:


Note that for $r(0)=P$ and $r(1)=Q$.

For the second:

Here you have a curve but the concept is the same, t=0 correspond to the strarting point and t=1 to the final one.

The parametrization can be obtained noting that a generic point in the curve must satisfy simultaneously:

$z=2-x^2-2y^2$ and $z=x^2$

thus, assuming x=t the coordinates of the point are


$z=x^2\implies z=t^2$

$z=2-x^2-2y^2\implies t^2=2-t^2-2y^2\implies y^2=1-t^2\implies y=\sqrt{1-t^2}$

Note that for y we have choose the positive value since it is requested that y goes from 1 to 0.

Note also that in this case the fact that t varies from 0 to 1 depends from the choice x=t with x varying from 0 to 1.

NOTE In both cases you are parametrizing the line (all the points of the lines) by a single real parameter t which varies form o to 1.


Hint: for the first one for $t=0\,$ you have the initial (which is $(0, 0, 0)$) and for $t=1$ you have the terminal point (which is $(3, 1, -2)$) of the line where you take the integral. Similarly, you do the same thing for the second integral.

  • $\begingroup$ Thanks I have understood intervals but I still have confusion about parametrization. Do I have to parametrize wrt point? I have parametrized wrt curves in 1st question I have obtained same r? $\endgroup$
    – usereb
    Dec 31 '17 at 13:16
  • $\begingroup$ @esrabasar: you can parametrize it in many ways and the range of $t$ changes according to your parametrization but it has to involve the initial and final points when put $t$ values. $\endgroup$
    – daulomb
    Dec 31 '17 at 13:46

"Points" were used for the parameterization. The question asked for the integral, along the given line, from (0, 0, 0) to (3, 1, -2). The reason "t" is between 0 and 1 in the parameterization r(t)=3ti+ tj- 2tk is that then r(0)= 0 and r(1)= 3i+ j- 2k- in other words, (0, 0, 0) and (3, 1, -2), the two endpoints of the line. As far as finding the parameterization itself, we have the two planes given by x- y+ z= 0 and x+ y+ 2z= 0. Any (x, y, z) on the line of intersection of the two planes must satisfy both. Subtracting the first equation from the second, we have 2y+ z= 0, z= -2y. Setting z= -2y in either equation, x= 3t. Using y itself "as parameter", that is, y= t, x= 3t and z= -2t. r(t)= xi+ yj+ kz= 3ti+ ti- 2tk.

For the "curve of intersection" of $z= 2- x^2- 2y^2$ and $z= x^2$, we have $2- x^2- 2y^2= x^2$ or $2x^2= 2- 2y^2$ so $x^2+ y^2= 1$, the circle with center at (0, 0) and radius 1. Since we only want the first quadrant, x and y are both positive and we go from (1, 0, 1) to (0, 1, 0) so $y= \sqrt{1- x^2}$. Taking x as parameter, x= t, $y= \sqrt{1- x^2}$, and $z= t^2$. When t= 0, that is x= 0, y= 1, z= 0 or (0, 1, 0). When t= 1, it is x= 1, y= 0, z= 1 or (1, 0, 1). t always lies between the values that give the beginning and ending points for the given path.

By the way, since it is also true that $cos^2(\theta)+ sin^2(\theta)= 1$ for all $\theta$, another "parameterization" for that curve would be $x= cos(\theta)$, $y= sin(\theta)$, $z= cos^2(\theta)$. And then the parameter, $\theta$ must be $0\le \theta\le \frac{\pi}{2}$ since, at $\theta= 0$, $x= cos(\theta)= 1$, $y= sin(\theta)= 0$, and $z= cos^2(\theta)= 1$, the endpoint (1, 0, 1), while at $\theta= \pi/2$., $x= cos(\theta)= 0$, $y= sin(\theta)= 1$, and $z= cos^2(\theta)= 0$, the other end point.


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