Answer:
V = (About) 22.2, Graph = First graph/Graph in the attachment
Step-by-step explanation:
Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.
![\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}](https://tex.z-dn.net/?f=%5Cmathrm%7BV%5C%3A%3D%5C%3A%5Cpi%20%5Cint%20_a%5Eb%5Cleft%28r%5Cright%29%5E2dy%5C%3A%7D%2C%5C%5C%5Cmathrm%7BV%5C%3A%3D%5C%3A%5Cint%20_1%5E3%5C%3A%5Cpi%20%5Cleft%5B%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2-1%5Cright%5Ddy%7D)
The plus 1 in '1 + 2/x' is shifting this graph up from where it is rotating, but the negative 1 is subtracting the area between the y-axis and the shaded region, so that when it's flipped around, it becomes a washer.
![V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\](https://tex.z-dn.net/?f=V%5C%3A%3D%5C%3A%5Cint%20_1%5E3%5C%3A%5Cpi%20%5Cleft%5B%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2-1%5Cright%5Ddy%2C%5C%5C%5C%5C%5Cmathrm%7BTake%5C%3Athe%5C%3Aconstant%5C%3Aout%7D%3A%5Cquad%20%5Cint%20a%5Ccdot%20f%5Cleft%28x%5Cright%29dx%3Da%5Ccdot%20%5Cint%20f%5Cleft%28x%5Cright%29dx%5C%5C%3D%5Cpi%20%5Ccdot%20%5Cint%20_1%5E3%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2-1dy%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3Athe%5C%3ASum%5C%3ARule%7D%3A%5Cquad%20%5Cint%20f%5Cleft%28x%5Cright%29%5Cpm%20g%5Cleft%28x%5Cright%29dx%3D%5Cint%20f%5Cleft%28x%5Cright%29dx%5Cpm%20%5Cint%20g%5Cleft%28x%5Cright%29dx%5C%5C%3D%20%5Cpi%20%5Cleft%28%5Cint%20_1%5E3%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2dy-%5Cint%20_1%5E31dy%5Cright%29%5C%5C%5C%5C)

Our exact solution will be V = π(4In(3) + 8/3). In decimal form it will be about 22.2 however. Try both solution if you like, but it would be better to use 22.2. Your graph will just be a plot under the curve y = 2/x, the first graph.
The size of the largest square is 36m². To find for the measures of the squares, use the common factors of 42 and 60. Among the common factors, choose the greatest.
Find the factors of 42 using the listing method.
42 - 1, 2, 3, 6, 7, 14, 21, 42
Find the factors of 60 using the listing method also.
60 - 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Identify the common factors of 42 and 60.
42 - 1, 2, 3, 6, 7, 14, 21, 42
60 - 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Common Factors: 1, 2, 3, 6
Therefore, the size of the largest square is 36m² since 6m x 6m is equals to 36m².
There are 7,920 minutes in 5.5 days
60minutes in an hour
24hours in a day
60 times 24 times 5.5 = 7,920