Minimizing Resultant Volume

The idea to solve this question is very similar to another recent problem .

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We want to minimize $\begin{aligned} V_{a,b} &=& \pi \int_0^1 y^2 \, dx \\ &=& \pi \int_0^1 (x^2 + ax + b)^2 \, dx \\ &=& \pi \int_0^1 (x^4 + x^3(2a) + x^2 (a^2 + 2b) + x(2ab) + b^2) \, dx \\ &=& \pi \left [\dfrac13 x^3(a^2 + 2b) +ab x^2 + \dfrac{ax^4}2 + b^2 x + \dfrac{x^5}5 \right]_{x=0}^{x=1} \\ &=& \pi \left [\dfrac13 x^3(a^2 + 2b) +ab x^2 + \dfrac{ax^4}2 + b^2 x + \dfrac{x^5}5 \right]_{x=0}^{x=1} \\ &=& \dfrac{\pi}{30} (10a^2 + 30ab + 15a + 30b^2 + 20b + 6) \end{aligned}$

What's left is to minimize the expression $10a^2 + 30ab + 15a + 30b^2 + 20b + 6$ via completing the square ,

$10a^2 + 30ab + 15a + 30b^2 + 20b + 6 = \dfrac58 (4a + 6b+3)^2 + \dfrac{15}2\left( b - \dfrac16 \right)^2 + \dfrac16 \geq 0 + 0 + \dfrac16 \; .$

Hence the minimum volume is $\dfrac{\pi}{30} \cdot \dfrac16 = \dfrac1{180} \pi$ , and it is attained when $\begin{cases} 4a + 6b + 3 = 0 \\ b - \dfrac16 = 0 \end{cases} \Leftrightarrow a = -1, b = \dfrac16$ .

Our answer is $1 + 180 = \boxed{181}$ .

Let ${\bf y = x^2 + ax + b }$ .

The volume of the region bounded by the curve ${\bf y = x^2 + ax + b }$ , the ${\bf x }$ axis, ${\bf x = 0, }$ and ${\bf x = 1 }$ , when revolved about the ${\bf x }$ axis is:

$$

${\bf V = \pi \int_{0}^{1} (x^2 + ax + b) ^2 \: dx = }$ $$

${\bf \pi \int_{0}^{1} (x^4 + 2ax^3 + (a^2 + 2b)x^2 +2abx + b^2 = }$ $$

${\bf \pi \int_{0}^{1} (\frac{1}{5}x^5 + \frac{a}{2}x^4 + (\frac{a^2 + 2b}{3})x^3 + abx^2 + b^2x) }$

${\bf \pi * (\frac{1}{5} + \frac{a}{2} + \frac{a^2 + 2b}{3} + ab + b^2) \implies }$ .

${\bf V(a,b) = \pi * (\frac{1}{5} + \frac{a}{2} + \frac{a^2 + 2b}{3} + ab + b^2) }$ $$

${\bf \implies \frac{\partial V}{\partial a} = \frac{1}{2} + \frac{2a}{3} + b = 0 }$ and ${\bf \frac{\partial V}{\partial b} = \frac{2}{3} + a + 2b = 0 \implies }$ $$

${\bf 4a + 6b = -3 }$ $$

${\bf 3a + 6b = -2 }$

${\bf \implies a = -1, b = \frac{1}{6} }$ $$

Let $D = [ \begin{array}{cc} \frac{\partial^2 V}{\partial a^2} & \frac{\partial^2 V}{\partial b \partial a} \\ \frac{\partial^2 V}{\partial a \partial b} & \frac{\partial^2 V}{\partial b^2} \\ \end{array} ]$ $$

If ${\bf det(D) > 0 }$ and ${\bf \frac{\partial^2 V}{\partial a^2} > 0}$ and ${\bf \frac{\partial^2 V}{\partial b^2} > 0, }$ then we have a relative min at ${\bf (a ,b) }$ $$

${\bf \frac{\partial^2 V}{\partial a^2} = \frac{2}{3} > 0 }$ and ${\bf \frac{\partial^2 V}{\partial b^2} = 1 > 0 }$ and ${\bf \frac{\partial^2 V}{\partial a \partial b} = \frac{\partial^2 V}{\partial b \partial a} = 0 }$ $$

${\bf \implies det(D) = \frac{2}{3} > 0 \implies }$ we have a relative min at ${\bf (a = -1,b = \frac{1}{6}) }$ and the volume ${\bf V = \frac{1}{180}\pi = \frac{a}{b}\pi \implies}$

${\bf a + b = 181. }$