Question

could someone help me with this

Answer 1

1) Symmetry. You need to calculate only one piece and multiply by 2. Right?

Answer 2

2) Since both are exactly the same size and shape, it also makes no difference which is cutting which.

Answer 3

3 & 4) It may be advantageous to restrict to the 1st Octant and multiply the result by 4, but I'm guessing the cyllindrical coordinates won't actually care about that.

Answer 4

Okay, that's enough talk. Let's see your initial attempt. You're probably close!

Answer 5

5) The area of a sphere or Radius 9 is \(4\pi\cdot 9^{2} ≈1018\) , so I would expect a final result in the neighborhood of 1/3 to 1/2 of that.

Answer 6

Note: I wouldn't NECESSARILY use cyllindrical coordinates. See what you think. Explore!

Answer 7

so we re to use sqrt(1+derivative of x^2+1+derivative of y ^2)
the derivative of y is 0

Answer 8

and the derivative of x is (2x)(x^2+81)^(-1/2)

Answer 9

i dont know how to integerate sqrt((5x^2+81)/(x^2+81)) and the boundaries in polar ot woiuld be from 0->2pi and 0->9 but this looks like it would turn into an ugly polar

Answer 10

i know x^2+81 simplifies into (x-9)(x+9)

Answer 11

but than what

Answer 12

We seem to be wandering off. Give this a good read and then let's get back to it.

http://mathworld.wolfram.com/SteinmetzSolid.html

Answer 13

k so i got an integral 9/sqrt(81-x^2)

Answer 14

is that right?

Answer 15

-sqrt(81-x^2)<y<sqrt(81-x^2) and -9<x<9 is that right??
im getting 324 but the answer is 648

Answer 16

Right. You have only the top half.

Answer 17

how do we know when we only have the top half and need to multiply by two?

Answer 18

How did you set it up? Did you consider the bottom half? This is my Point #1 at the very top.

Answer 19

oh see i didnt understand that
i simply took the x derivative of the cylinder and used the circle equations to find the limits of integration

Answer 20

And since youconsidered only positive z, youmanaged only the top half.

Let's see how this compares to my intuition: \(\dfrac{648}{1018} = 0.637\) -- Well, that's quite a bit larger than my rough estimate (1/3 to 1/2), but it is the same order of magnitude.

Answer 21

so you need to be able to see the 3d graph inorder to resolve these sort of problems

Answer 22

It does help, but calculus has application far beyond 3D. If you ALWAYS rely on that visualization, you will continue to struggle. If youfollow your work carefully, ther are clues that do not come from a picture:

Given \(x^{2} + z^{2} = 81\)

Solve for z: \(z = \pm\sqrt{81-x^2}\)

There is your clue. We might say to ourselves at this point, "Oh, that's annoying. Let's stay above the x-y plane and just remember that we're getting only half the result." This allows us to use "+" only and ignore the "-".

Answer 23

ohhh thanks i see it!!!