Question

I am currently stuck on a calc two problem. The problem is attached in the image below. I thank you for any help or assistance you can provide me with this problem.

Answer 1

assuming you want to find the volume as it gets rotated like this
http://prntscr.com/62m08h
then...
find x in terms of y
so like y = (1/9)x^2 solve for x
I got like x = sqrt(9y)

the at x = 8, y = 64/9

then we use the disc formula
which is something like
volume = pi integral from 0 to 64/9 of ((sqrt(9y))^2 - (0)^2) dy

which should get you your volume... (I'm not entirely sure sorry)

volume would be in ft^3
and to get weight you would multiply the number you got by 62.4 to get your answer

Answer 2

@iambatman @Data_LG2 I'm not sure if I helped that right xD
(cuz I thought I knew it...then I got a 55% on that test last week)

Answer 3

I think the work is the amount of water lifted to the top of the parabola.
At each height y, the very thin "cylinder" with height dy and radius r(y) (i.e. radius is a function of the height) has volume
\[ 2 \pi \ r(y)^2 \ dy \ ft^3\]
and the weight is
\[ 62.4 \ \frac{lbs}{ft^3} \cdot 2 \pi \ r(y)^2 \ dy \ ft^3 \]
This weight must be raised from y up to the top at y= 64/9, which is a distance (64/9-y)
that is, the work to raise the water in the thin disk is
\[ 62.4 \ \frac{lbs}{ft^3} \cdot 2 \pi \ r(y)^2 \ dy \ ft^3 \cdot \left(\frac{64}{9}-y\right) \ ft\]
using \( r(y)^2= x^2 \) and \( x^2= 9y\)
\[ 62.4 \cdot 2 \pi \cdot 9y \left(\frac{64}{9}-y\right) \ dy \ ft-lb\]
integrate over y to find the total work
\[ 124.8 \pi \int_0^{\frac{64}{9}} 9y \left(\frac{64}{9}-y\right) \ dy \ ft-lb\]

Answer 4

I appreciate the help. I managed to solve the problem with a help of a friend. The answer given by phi was close but not fully correct. Thank you.

Answer 5

oops, I wrongly used area of a circle= \( 2 \pi r^2 \), conflating area and circumference formulas.
It should be \( A= \pi r^2\)
so the posted answer has an extra factor of 2 which should be removed.