Math 3 Problem: Determine whether the following statements are true and give an

Question

Math 3 Problem:

Determine whether the following statements are true and give an explanation or counterexample.

a) Let R be the unit disk centered at (0,0). Then

b) The average distance between the points of the hemisphere

and the origin is 2 (no integral needed).

c) The integral

is easier to evaluate in polar coordinates than in Cartesian coordinates.

The back of the book says the answer for part a is False and for parts b and c are true. Can someone post the explanation or counterexample for each of these three parts?

Math 3 Problem: Determine whether the following statements are true and give an explanation or counterexample. a) Let R be the unit disk centered at (0,0). Then e^x^2^y^2 dx dy is easier to evaluate in polar coordinates than in Cartesian coordinates. The back of the book says the answer for part a is False and for parts b and c are true. Can someone post the explanation or counterexample for each of these three parts? Double integrate z = sqrt(4-x^2-y^2) and the origin is 2 (no integral needed). c) The integral double integrate {R} (x^2+y^2) dA = double integrate r^2 dr dtheta b) The average distance between the points of the hemisphere

Explanation / Answer

Q1 Its False

Because any point on disk be (x,y) where writing in polar coordinates x=r*cost y=r*sint

x^2+y^2=r^2 & dx dy get converted to dx=-rSint dt dy=rcost dt hence integral becomes (r^4 *sint cost dr dt)

Q2 True

s^2+y^2+z^2=4

Radius =2 distance of any point on the hemisphere from the origin is Radius ie 2

Hence the average distance must be also 2.

3. The integral is indeed simpler to compute in polar than in cartesian

Take x=r*cost y=r*sint

Rplace y by 1-x^2 =1-rcost^2

Not its just exp(rcost^2 *(1-rcost^2) which is simpler to solve.

The limit 1-y^2 changes to (sint)^2

In cartesian the equation is almost impossible to solve

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