Calculate unknown variable when surface area is given. (Calculus) I hope I could get some help with the following calculus problem. I literally have no idea how to tackle this.

Here it is:
For every value of p the following function is given:
$$f_{p} (x) = p(x^2-1)$$
Also:
$$V_{p}$$
is the area contained by:
$$f_{p} \,\,\,and\,\,\,\, f_{1}$$
Here's the problem I need to solve:
Calculate p exactly when surface area Vp equals 4 is given.

No idea how to tackle this because no lower limit or upper limit is given. I haven't seen this before using an indefinite integral.
Your help is very much appreciated,
Bowser
*Calculate exactly meaning without use of a calculator.
 A: You can calculate explicitly the intersection of the curves $y = p(x^2-1)$ and $y = (x^2-1)$ for all values of $p$ and you get $x_1=-1$ and $x_2=1$.
Then you can calculate the area between the two curves to get the value of p solving (if $p \le 1$):
$$4 = \int_{-1}^1 (p(x^2 -1) - (x^2 -1) )dx = \int_{-1}^1 (p-1)(x^2 -1)  dx $$
If $p \ge 1$ you have to change the sign of the integral because then $(x^2-1) \ge p(x^2 -1 )$ in the interval where you want to integrate.
A: $$f_1=x^2-1\;,\;\;f_p=p(x^2-1)$$
The intersection of both functions above:
$$x^2-1=p(x^2-1)\iff(p-1)(x^2-1)=0\iff x=\pm1$$
and these are your integration bounds. Also
$$f_p\ge f_1\iff(p-1)(x^2-1)\ge0\iff p-1\le 0\;,\;\;\text{since}\;\forall x\in[-1,1],\;\;x^2-1\le 0 $$
so if $\;p\ge1\;$ , then always $\;f_1\ge f_p\;$ , and then
$$4=V_p=\int_{-1}^1\left[(x^2-1)-p(x^2-1)\right]dx=2(1-p)\int_0^1(x^2-1)dx=$$
$$=2(1-p)\left(\frac13-1\right)=\frac43(p-1)\implies p=4$$
If $\;p\le1\;$ you have to reverse the inequalities above (Warning: in this case the solution IS NOT $\;-4\;$. Check it) 
A: The area between $f_1$ and $f_p$ is 
\begin{align}
&\int_{-1}^1|f_1(x)-f_p(x)|d x\\
&=\int_{-1}^1|(1-p)(x^2-1)|d x\\
&=|1-p|\int_{-1}^1( 1-x^2) d x \\
&=\frac43 |1-p| \\
&=4
\end{align}
thus, $p=4\text{ or }p=-2$
