Integral Calculus Including Differential Equations -
Kael nodded grimly. "That’s the energy. If you release a counter-vortex with exactly that integrated strength, shaped like ( u(r) = 48 - \frac{3}{4}r^3 ), the sum of the two integrals will be zero. The Churnheart will still itself."
[ P = \int_{0}^{R} v(r) , dr = \int_{0}^{4} \frac{3}{4} r^3 , dr ]
[ r v = \int 3r^3 , dr = \frac{3}{4} r^4 + C ] Integral calculus including differential equations
[ \mu(r) = e^{\int \frac{1}{r} dr} = e^{\ln r} = r ]
[ \int_{0}^{4} \frac{3}{4} r^3 , dr = \frac{3}{4} \cdot \left[ \frac{r^4}{4} \right]_{0}^{4} = \frac{3}{16} \left( 4^4 - 0 \right) ] Kael nodded grimly
The integrating factor ( \mu(r) ) was:
She multiplied through:
[ 4^4 = 256, \quad \frac{3}{16} \times 256 = 3 \times 16 = 48 ]
Integrating both sides with respect to ( r ): The Churnheart will still itself
[ \frac{dv}{dr} + \frac{1}{r} v = 3r^2 ]
[ v(r) = \frac{3}{4} r^3 ]
