# Math Help - Intersection of paraboloid and plane

1. ## Intersection of paraboloid and plane

If C is the curve in reals^3 which is the intersection of the paraboloid {(x,y,z) s.t. z= (x^2) + (y^2)}
With the plane {(x,y,z) s.t. z= 3 -2y}
And L is the segment of C which lies in the half-space x>or=0. What is integral fds with bottom limit L (on integral) where f(x,y,z) := 5xy +xz?

2. Originally Posted by maximus101
If C is the curve in reals^3 which is the intersection of the paraboloid {(x,y,z) s.t. z= (x^2) + (y^2)}
With the plane {(x,y,z) s.t. z= 3 -2y}
And L is the segment of C which lies in the half-space x>or=0. What is integral fds with bottom limit L (on integral) where f(x,y,z) := 5xy +xz?
If I am reading this correctly then the first thing you need to find is C.
IF you subract the equation of the plane from the equation of the paraboloid we get

$\begin{array}{rrrr} z=&x^2&+y^2& \\ -z=&& +2y&-3 \\ 0=& x^2&+y^2+2y&-3\end{array}$
Or
$x^2+y^2+2y=3 \iff x^2+(y+1)^2=4$

This is a circle of radius $2$ in the xy plane and can be parametrized as $x=2\cos(t)$ and $y=-1+2\sin(t)$

Now if we put this into the equation of the plane we get

$z=3-2y=3-2(-1+2\sin(t))=5-4\sin(t)$

So now we have a parametrization of the needed curve
$\vec{r}(t)=[2\cos(t)]\vec{i}+[-1+2\sin(t)]\vec{j}+[5-4\sin(t)]\vec{k}
$

Can you finish from here?

3. thank you that helped a lot, so I found f(x,y,z) = 5xy+xz= 12cos(t)sin(t) and ds=2sqrt(1+4(cost)^2) dt

so what are the limits for my integral? in finding ∫ [L] f(x,y,z)ds and could you explain how we found them.

4. Originally Posted by maximus101
And L is the segment of C which lies in the half-space x>or=0. [/SIZE][/FONT]
so what are the limits for my integral?
Well $x=2\cos(t)$ so $2\cos(t) \ge 0 \implies ...$

5. so bounds are -pi/2 lower and pi/2 upper

6. Yes

7. cool thank you