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Multiple Intergration, Area, Center of Mass, Centroid and Jacobian

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1.Given the region R bounded by y=2x+2 , 2y=x and 4.
a) Set up a double integral for finding the area of R.
b) Set up a double integral to find the volume of the solid above R but below the surface
f(x,y) 2+4x.
c) Setup a triple integral to find the volume of the solid above R but below the surface f(x,y)=-x^2 +4x.
d) Set up the integral to find the moment of the solid in part b) about the xy-plane.
e) Set up the integral for finding the surface area of f(x, y) = ?x^2 + 4x above R.
1) Find the mass of the solid from part c) if d(x,y,z) = 2x.
g) If we assume a lamina with the shape of R is of homogeneous density, flnd the centroid.

2. Given the solid bounded by the two spheres x2 + y2 + z2 =1 and x2 + y2 + z2 =9 and the upper nappe of the cone = 3(x2 + y2),
a) Set up the integral for finding the volume using cylindrical coordinates.
b) Set up the integral for finding the volume using spherical coordinates.

3. Given the integral below, use u and v substitution to change the variables. Assume that the the integral is to be evaluated over the region R bounded by x = 2y, y = 2x, x + y =1, and x + y =2. (Do not evaluate the integral)

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Probems involving Multiple Intergration, Area, Center of Mass, Centroid, Moment, Surface Area and Jacobian are solved. The solution is detailed and well presented. The response received a rating of "5" from the student who originally posted the question.

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Solution to question 1.

a) The area S of R is

b) The volume V is given by

c) Denote the region by . Then we have

d) The moment is given by

e) The surface area S* is given by

f) The mass of solid from part c) is

g) Since there are many homogenous ...

Solution provided by:
  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
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  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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