# Thread: Just need a little bit of help with these Calc questions-- integrals and volume

1. ## Just need a little bit of help with these Calc questions-- integrals and volume

1) At a processing plant, over the course of an 8-hour day, workers move material into a pile, which is removed at a constant rate by a conveyor belt. The workers move material into the pile at rate 200e−0.5t units/hr while the conveyor belt removes the material at 50 units/hr.

(a) Find the net change in the size of the pile over the first 2 hours, and over the first 8 hours. Round your answers to the
nearest hundredth.

2 hours =
8 hours =

(b) At what time is the amount of material in the pile the largest and the smallest?

largest t=
smallest t=

2)
Consider the region in the first quadrant bounded by the line 8xy = 0 and the curve 8x^3y = 0.

Find the volume of the solid of revolution generated by revolving this region about the line x = −1
using the disk/washer method.

Thanks!!!!

2. ## Re: Just need a little bit of help with these Calc questions-- integrals and volume

The net change at some time $\Delta(t)$ is the amount added minus the amount removed between time 0 and time $t$

$d(t) = 200e^{-0.5t}-50$

$\Delta(t) = \displaystyle \int_0^t d(\tau)~d\tau$

$\Delta(2) = \displaystyle \int_0^2 d(t)~dt$

$\Delta(8) = \displaystyle \int_0^8 d(t)~dt$

To solve for the maximum amount find $t$ such that

$\dfrac{d}{dt}\Delta(t)=0$

It should be clear that eventually the constant removal rate will become larger than the addition rate
and thus at some time the amount of material will be 0. This is the minimum amount of material.

Solve for $t$ such that $\Delta(t) = 0$

3. ## Re: Just need a little bit of help with these Calc questions-- integrals and volume

2) Consider the region in the first quadrant bounded by the line 8x − y = 0 and the curve 8x^3 − y = 0.

Find the volume of the solid of revolution generated by revolving this region about the line x = −1
using the disk/washer method.
washers about the line $x=-1$

$R(y) = \dfrac{\sqrt[3]{y}}{2} + 1$

$r(y) = \dfrac{y}{8} + 1$

$\displaystyle V = \pi \int_0^8 [R(y)]^2 - [r(y)]^2 \, dy$

Just an aside, might be easier to calculate the volume using concentric shells w/r to $x$