<s>[INST] finding the value of $a$ for minimal solid Given that $y=x^2+ax-0.25$ is bounded by $x=0$ and $y=1$ and spinning around the x-axis.
What is the value of $a$ for min volume of that solid?
My attempt: $x^2+ax-0.25=1$ so $x=\frac{-a\pm \sqrt{a^2+5}}{2}$.
Then, $f(a)=\pi \int_{0}^{\frac{-a+\sqrt{a^2+5}}{2}}(1-(x^2+ax-0.25)^2)dx$
But that will lead to: $f(a)=1/480 (-225a+20a^3+8a^5+200 \sqrt{5+a^2}-8a^4 \sqrt{5+a^2})π$
And then $f'(a)=0$ will be very hard to solve.
Any ideas?
Update: wolfram gave me an answer but i wonder if it can be done without that.
Thanks.
 [/INST] I think you overcomplicated the problem.
You only need to find out what is the minimum of :
$f(a)=\int_{0}^{\frac{-a+\sqrt{a^2+5}}{2}}|x^2+ax-0.25|dx$
Since the volume is going to be the rotation of this area around the axis $x$. If you minimize the area, you minimize the volume. 
The function $y=x^2+ax-0.25$ has a positive zero in $x=\dfrac{-a+\sqrt{a^2+1}}{2}$, which is smaller than $\dfrac{-a+\sqrt{a^2+5}}{2}$, so that divide the integral in two domains.
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