The solid shown in the figure below consists of a cylinder of the radius (r) and height (h) and a hemispherical void of the radius (r). The dimensions at a given instant (t) are: $$h(t)=3t^2+2;r(t)=8-\frac{t^2}{4}$$ Find the rate of change of the volume (V) and surface area (S) of the solid at $t = 4$ seconds. State whether (V) and (S) are increasing, decreasing or neither. Note:
The volume of Sphere =$\frac{4}{3}πr^3$
$(Ans: \frac{dV}{dt} = - 1105.84 (Deceasing))$
I tried this: $$\frac{dV}{dt}=\frac{\partial{V}}{\partial{r}}*\frac{dr}{dt}$$
and said: $$t=4$$ $$r(4)=4$$ $$\frac{\partial{V}}{\partial{r}}=4πr^2;\frac{dr}{dt}=\frac{-t}{2}$$
so:$$\frac{dV}{dt}=(4πr^2)*(\frac{-t}{2})$$
and the final result I get:
$$\frac{dV}{dt}=128π$$
What am I doing wrong?
Volume--> $$V=\pi r^2h-\frac{2}{3}\pi r^3$$ $$V=\pi (8-\frac{t^2}{4})^2(3t^2+2)-\frac{2}{3}\pi (8-\frac{t^2}{4})^3$$ $$\frac{dV}{dt}=\pi[(3t^2+2)*2(8-\frac{t^2}{4})*(\frac{-t}{2})+(8-\frac{t^2}{4})^2*6t]-[\frac{2}{3}\pi *3(8-\frac{t^2}{4})^2*(\frac{-t}{2})]$$ $$\frac{dV}{dt}=\pi[\frac{19}{16}t^5-\frac{103}{2}t^3+432t]$$
At $t=4s$, $$\frac{dV}{dt}=\pi[\frac{19}{16}*4^5-\frac{103}{2}*4^3+432*4]=-352\pi\approx-1105.84$$
Surface Area--> $$A=2\pi rh+3\pi r^2$$ $$A=2\pi(8-\frac{t^2}{4})(3t^2+2)+3\pi(8-\frac{t^2}{4})^2$$ $$\frac{dA}{dt}=\pi[2(3t^2+2)(\frac{-t}{2})+3*2(8-\frac{t^2}{4})(\frac{-t}{2})]$$ $$\frac{dA}{dt}=\pi[\frac{-9}{4}t^3-26t]$$
At $t=4s$, $$\frac{dA}{dt}=\pi[\frac{-9}{4}*4^3-26*4]=-248\pi\approx-779.115$$