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.

Solutions Collecting From Web of "Calculate unknown variable when surface area is given. (Calculus)"

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.

$$f_1=x^2-1\;,\;\;f_p=p(x^2-1)$$

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)

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, $p=4\text{ or }p=-2$