Answer:
Step-by-step explanation:
Consider the graphs of the 
and 
.
By equating the expressions, the intersection points of the graphs can be found and in this way delimit the area that will rotate around the Y axis.
then 
o 
. Therefore the integration limits are:
and 
The inverse functions are given by:
and 
. Then
The volume of the solid of revolution is given by:
![\int\limits^{64}_ {0} \, [2\sqrt{y} - \frac{y}{4}]^{2} dy = \int\limits^{64}_ {0} \, [4y - y^{3/2} + \frac{y^{2}}{16} ]\ dy = [2y^{2} - \frac{2}{5}y^{5/2} + \frac{y^{3}}{48} ]\limits^{64}_ {0} = 546.133 u^{2}](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B64%7D_%20%7B0%7D%20%5C%2C%20%5B2%5Csqrt%7By%7D%20-%20%5Cfrac%7By%7D%7B4%7D%5D%5E%7B2%7D%20%20dy%20%3D%20%5Cint%5Climits%5E%7B64%7D_%20%7B0%7D%20%5C%2C%20%5B4y%20-%20y%5E%7B3%2F2%7D%20%2B%20%5Cfrac%7By%5E%7B2%7D%7D%7B16%7D%20%5D%5C%20%20dy%20%3D%20%5B2y%5E%7B2%7D%20-%20%5Cfrac%7B2%7D%7B5%7Dy%5E%7B5%2F2%7D%20%2B%20%5Cfrac%7By%5E%7B3%7D%7D%7B48%7D%20%5D%5Climits%5E%7B64%7D_%20%7B0%7D%20%3D%20546.133%20u%5E%7B2%7D)
No because 2/3 to 3/5 and 3/5 to 5/7 arent proportional.
For volume you would do 20 x 8.5 x 7.5 = 1275 ft^3
For surface are you would add up all the area of the faces
left and right face = 8.5 x 7.5 = 63.75 each
front and back = 20 x 8.5 = 170 each
top and bottom = 20 x 7.5 = 150 each
total = 767.5 ft^2
