1. ## area of region

1. Sketch the region enclosed by the curves and find its area.
 = ,  = 4,  = − + 2
2. Find the volume of the solid that results when the region enclosed by the given curves is
 =
cos,  =

4
,  =

2
,  = 0
3. Solve the initial value problem by separation of variables


=
4
 + cos
, 1 = 
4. Solve the initial value problem by the method of integrating factors


− = , 0 = 3
5. Find the equation of the plane that contains the points 1,−2,0, 3,1,4 and 0,−1,2.

2. Originally Posted by res88
...
5. Find the equation of the plane that contains the points 1,−2,0, 3,1,4 and 0,−1,2.

to #5:

I assume that you mean P(1, -2, 0), Q(3, 1, 4) and R(0, -1, 2).

You need a fixed point (I take P) and 2 direction vectors:

$\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}= \langle 2, 3, 4\rangle$

$\overrightarrow{PR}=\overrightarrow{OR}-\overrightarrow{OP}= \langle -1, 1, 2\rangle$

Then the equation of the plane is:

$\vec v = \langle1,-2,0 \rangle+s \cdot \langle 2, 3, 4\rangle + t \cdot \langle -1, 1, 2\rangle$

As a parametric system of equations:
$\begin{array}{l}x=1+2s-t \\ y=-2+3s+t \\ z=4s+2t\end{array}$

The normal vector of the plane is:

$\langle 2, 3, 4\rangle \times \langle -1, 1, 2\rangle=\langle 2, -8, 5\rangle$

And therefore the equation of the plane in normal form (keep in mind that the point P belongs to the plane!):

$2x-8y+5z-18=0$