Question

The base of a solid is the circle
x^2 + y^2 =16. Find the volume of the solid given that the cross
sections perpendicular to the y − axis are squares.

Answer 1

How do I find my bounds / limits here?

Answer 2

@fwizbang @agent0smith

Answer 3

i know r=4 and the center is at the orgin

Answer 4

Is this meant to be a single double or triple integration?(i.e. what class are you in?)

Answer 5

calc 2 single int

Answer 6

OK. Since we know the cross section perpendicular to the y-axis,
y is the variable to integrate. So, for a given y, what's the biggest and smallest x can get?

Answer 7

4

Answer 8

If y is 3, then x can't be 4 if x^2+y^2 < 16.....

Answer 9

Assume y is fixed and solve for x.

Answer 10

\[ysqrt{16-x^2}\]

Answer 11

Sorry having some technical difficulties

Answer 12

Good(I think). Remember that there's a positive and negative root so that\[-\sqrt{16-y^2}<x<sqrt{16-y^2\]. So this gives us the length of one side of the square.

Answer 13

\[y < \sqrt{16-x^2}\]

Answer 14

We're going to integrate y, so we want to solve for x.

Answer 15

Its just the same though isnt it?

Answer 16

In terms of the circle, yes. But the square cross sections are in the x-z plane,perpendicular to the y-axis, so it'll be easier to integrate over y.

Answer 17

So visually this is a tipped over cylinder with a bunch of squares stacked inside in three dimensions

Answer 18

How would you solve for x?

Answer 19

Sorry, I had a hard time grasping this concept in class too.

Answer 20

Since the edge of the circle is x^2+y^2=16, the interior is

\[-\sqrt{16-y^2}<x< \sqrt{16-y^2}\]

Answer 21

I guess I visualized this with the y-axis pointing up and the x-z plane horizontal.
Then the Hemisphere is essentially a stack of square plates.

Answer 22

So assuming y is fixed, what is the length of the square edge?

Answer 23

im not sure