Estimate the volume (quick Calc, 10 points!)?

Question

(You don't need to make models or cut anything, it is non-physical math :)

"The nose "cone" of a rocket is a paraboloid obtained by revolving the curve:
√x, 0 ≤ x ≤ 5 about the x axis where x is measured in feet. Estimate the volume of the nose cone by partitioning [0,5] into 5 subintervals of equal length, slicing the cone with planes perpendicular to the x-axis at the subintervals' midpoints, constructing cylinders of height 1 based on cross sections at these points, and finding the volumes of these cylinders."

Simply, estimate the volume of the nose cone.

I will give out best answer points at will, thank you!

Answer 1

Since you are on the interval [0,5] the 5 equal sub-intervals are:



0-1, 1-2, 2-3, 3-4, and 4-5. The mid points of each of these intervals is:

.5, 1.5, 2.5, 3.5, and 4.5. Since each interval is 1 unit in height/length, you have one piece of the puzzle. You are making cylinders and to find the volume of a cylinder you need the height (which we know) and the radius. The radius, for this problem, is the distance between the x-axis and the the curve √x at each of the midpoints. In other words the radius of each cylinder, where m is the midpoint of the interval, is √m. The radius for each cylinder is √.5, √1.5, √2.5, √3.5, and √4.5. The volume of a cylinder is πhr^2. For this problem, h = 1. Now all you have to do is sum the 5 volumes:



π(√.5)^2*1 + π(√1.5)^2*1 + π(√2.5)^2*1 + π(√3.5)^2*1 + π(√4.5)^2*1



This turns into:



π*.5*1 + π*1.5*1 + π*2.5*1 + π*3.5*1 + π*4.5*1



Factor out the π and get:



π(.5 + 1.5 + 2.5 + 3.5 + 4.5) = 12.5π cubic units

Answer 2

Your summation is of 5 terms where the domain of f(x) will be {.5, 1.5, 2.5, 3.5, 4.5} so the

area of the circular part is π [f(x)]² and the volume of each (of 5 cylinders, whose axis is the

x-axis) will π [f(x)]² [1].



We get a volume of 12.5π ≈ 39.26991 (which happens to be the exact volume by integration).

Answer 3

simply ,

Interval 1 , [0,1] , Dx= (1-0) =1 , midpoint t= (1+0)/2 =1/2 , Radius of revolution R= f(t) = sqrt (1/2)=0.707 , volume =V1= piR^2 Dx = pi (1/2)*1 =pi/2



Since we need for piR^2 , R^2 , y^2 = R^2 = t , so



I2 , [1,2] , Dx=1 , t=( 1+2)/2 = 3/2 R^2=(3/2) V2=pi(3/2) = 3pi/2



I3 ,[2,3] , Dx=1 , t=(2+3)/2 = 5/2 R^2=5/2 , V3= 5pi/2



I4 [3,4] , Dx=1 ,t=(3+4)/2= 7/2 R^2=7/2 , V4= 7pi/2



I5 [4,5] , Dx=1 t= 9/2 R^2=9/2 V5= 9pi/2



V Total ~ pi/2+3pi/2+5pi/2+7pi/2+9pi/2 = 25pi/2 ~ 39.27 (u3)

Answer 4

∑ π(f(x_i))^2 * ∆x

where f(x_i) = √(x_i) , ∆x = 1

∑ π(√(x_i))^2 * 1

= ∑ π(x_i)

= π * ∑ x_i

x_i actually won't be x_i, but will be (x_i + x_(i-1)) / 2 because it's the midpoint of the subintervals.

π * ∑ (x_i + x_(i-1)) / 2

= (π/2) * ∑ x_i + x_(i-1)

and you're summing this up from i = 1 to 6.

(π/2) * [ (1 + 0) + (2 + 1) + (3 + 2) + (4 + 3) + (5 + 4) + (6 + 5) ]

= (π/2) * [ 1 + 3 + 5 + 7 + 9 + 11 ]

= (π/2) * 36

= 18π

≈ 56.5487

Answer 5

actually no longer A. Can gasoline be as mild/heavy as a stable textile? do no longer think of so. No no no no longer B. volume of substance "interior" some thing??? How with reference to the substance itself? Why incorporate in user-friendly terms the interior? How with reference to the substance it self? :) C, do no longer think of so. Gasses have volumes. Gasses have not got "compacted" molecules. it fairly is a D for me ,y expensive. :o) unsure nevertheless. it relatively is why I gave my motives. desire I helped.