I need some to report my last 4 questions
1 answer:
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Answer:
first exercise
V = 16 π
second exercise
V= 68π/15
Step-by-step explanation:
Initially, we have to plot the graph x = (y − 5)2 rotating around y = 3 and the limitation x = 4
<em>vide</em> picture 1
The rotation of x = (y − 5)2 intersecting the plane xy results in two graphs, which are represented by the graphs red and blue. The blue is function x = (y − 5)2. The red is the rotated cross section around y=3 of the previous graph. Naturally, the distance of "y" values of the rotated equation is the diameter of the rotation around y=3 and, by consequence, this new red equation is defined by x = (y − 1)2.
Now, we have two equations.
x = (y − 5)2
xm = (y − 1)2 (Rotated graph in red on the figure)
The volume limited by the two functions in the 1 → 5 interval on y axis represents a volume which has to be excluded from the volume of the 5 → 7 on y axis interval integration.
Having said that, we have two volumes to calculate, the volume to be excluded (Ve) and the volume of the interval 5 → 7 called as V. The difference of V - Ve is equal to the total volume Vt.
(1) Vt = V - Ve
Before start the calculation, we have to take in consideration that the volume of a cylindrical shell is defined by:
f(y) represents the radius of the infinitesimal cylinder.
Replacing (2) in (1), we have
V = 16 π
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Second part
Initially, we have to plot the graph y = x2 and x = y2, the area intersected by both is rotated around y = −7. On the second image you can find the representation.
<em>vide</em> picture 2
As the previous exercise, the exclusion zone volume and the volume to be considered will be defined by the interval from x=0 and y=0, to the intersection of this two equations, when x=1 and y = 1.
The interval integration of equation y = x2 will define the exclusion zone. By the other hand, the same interval on the equation x=y2 will be considered.
Before start the calculation, we have to take in consideration that the volume of a cylindrical shell is defined by:
Notice that in the equation above, x and y are switched to facilitate the calculation. f(x) is the radius of the infinitesimal cylinder
Having this in mind, the infinitesimal radius of equation (3) is defined by f(x) + radius of the revolution, which is 7. The volume seeked is the volume defined by the y = x2 minus the volume defined by x=y2. As follows:
V= 68π/15
