Find volume of circle. Cross-sections are squares. What am I doing wrong? :/

• Lo.Lee.Ta.
In summary, the first problem has the incorrect answer of 36, while the second problem gets the correct answer of 576.
Lo.Lee.Ta.
Hi, everyone. I am just trying to do some practice problems on finding volume.

So this is the one I'm working on:

1. "Find the volume of the solid described below:

The solid lies between the planes perpendicular to the x-axis at x=6 and x=-6. The cross-sections perpendicular to the axis on the interval -6≤x≤6 are squares whose bases run from the semicircle y= -√(36 - x^2) to the semicircle y= √(36 - x^2)."2. I first drew out the object, which is a sphere with a radius of 6.

Every cross-section is a square, so each will have an area s^2. The area of each cross-section is 2(√(36 - x^2))^2.

To find the volume, the dx represents the thickness of each square that fills the sphere.

2∫-6 to 6 of (√(36 - x^2))^2

= 2∫36 - x^2

= 2(36x - (x^3)/3)dx |-6 to 6

= Plug in values... Answer = 576units^3

But this is counted as the wrong answer! What am I doing wrong here?! :/

Lo.Lee.Ta. said:
Hi, everyone. I am just trying to do some practice problems on finding volume.

So this is the one I'm working on:

1. "Find the volume of the solid described below:

The solid lies between the planes perpendicular to the x-axis at x=6 and x=-6. The cross-sections perpendicular to the axis on the interval -6≤x≤6 are squares whose bases run from the semicircle y= -√(36 - x^2) to the semicircle y= √(36 - x^2)."

2. I first drew out the object, which is a sphere with a radius of 6.

Every cross-section is a square, so each will have an area s^2. The area of each cross-section is 2(√(36 - x^2))^2.

To find the volume, the dx represents the thickness of each square that fills the sphere.

2∫-6 to 6 of (√(36 - x^2))^2

= 2∫36 - x^2

= 2(36x - (x^3)/3)dx |-6 to 6

= Plug in values... Answer = 576units^3

But this is counted as the wrong answer! What am I doing wrong here?! :/

I think the area of each cross section should be (2√(36 - x^2))^2 not 2(√(36 - x^2))^2. See the difference?

Oh, yes, I see what you mean. Since every cross-section has a side of 2√(36 - x^2), this whole thing is what needs to be squared.

But- I did this in another problem, and it led me to the wrong answer! That is why I switched it to 2(√(36 - x^2))^2.

"Find the volume of the solid describe below:

The solid lies between the planes perpendicular to the x-axis at x=0 and x=6 . The cross-sections perpendicular to the axis on the interval 0≤x≤6 are squares whose diagonals run from the parabola y= -√(x) to the parabola y= √(x)."

So when I drew out the graph, it looked sort of like a parabola turned on its side.

Since every cross-section is a square, the area = s^2.
Following the should-be-correct-method, the area of each square = (2√(x))^2

∫0 to 6 of (2√(x))^2dx

= ∫4xdx

= 4(1/2*x^2) |0 to 6

= 4(1/2(6)^2) - 0

= 72
This is counted as the wrong answer! The correct answer is 36!

I get 36 only when I separate the 2 from the (√(x))^2. But we already went over how that would be incorrect.

So what is wrong here...?
Thanks so much for the help! :)

Lo.Lee.Ta. said:
Oh, yes, I see what you mean. Since every cross-section has a side of 2√(36 - x^2), this whole thing is what needs to be squared.

But- I did this in another problem, and it led me to the wrong answer! That is why I switched it to 2(√(36 - x^2))^2.

"Find the volume of the solid describe below:

The solid lies between the planes perpendicular to the x-axis at x=0 and x=6 . The cross-sections perpendicular to the axis on the interval 0≤x≤6 are squares whose diagonals run from the parabola y= -√(x) to the parabola y= √(x)."

So when I drew out the graph, it looked sort of like a parabola turned on its side.

Since every cross-section is a square, the area = s^2.
Following the should-be-correct-method, the area of each square = (2√(x))^2

∫0 to 6 of (2√(x))^2dx

= ∫4xdx

= 4(1/2*x^2) |0 to 6

= 4(1/2(6)^2) - 0

= 72
This is counted as the wrong answer! The correct answer is 36!

I get 36 only when I separate the 2 from the (√(x))^2. But we already went over how that would be incorrect.

So what is wrong here...?
Thanks so much for the help! :)

In the first case they are giving you the side of the square s. Then the area is s^2. In the second case they are giving you the diagonal of the square. If the diagonal of the square is s, what's the area?

Oh, right! Thanks!

I thought that if the diagonals run from the parabola y=-√(x) to y=√(x),
then each diagonal should equal 2*√(x)

So since eace DIAGONAL = 2√(x), we have to figure out the length of each SIDE!

2√(x)= side*√2

[2√(x)]/√(2) = side

∫0 to 6 of [(2√(x))/√(2)]^2 dx

= ∫4x/2 dx= ∫2x dx

= (2x^2)/2 |0 to 6

= (6)^2 - (0)^2

= 36

So, yay, I get the right answer. But am I thinking about the diagonals and sides correctly?

Thank you! :)

Lo.Lee.Ta. said:
Oh, right! Thanks!

I thought that if the diagonals run from the parabola y=-√(x) to y=√(x),
then each diagonal should equal 2*√(x)

So since eace DIAGONAL = 2√(x), we have to figure out the length of each SIDE!

2√(x)= side*√2

[2√(x)]/√(2) = side

∫0 to 6 of [(2√(x))/√(2)]^2 dx

= ∫4x/2 dx= ∫2x dx

= (2x^2)/2 |0 to 6

= (6)^2 - (0)^2

= 36

So, yay, I get the right answer. But am I thinking about the diagonals and sides correctly?

Thank you! :)

Yes you are. If they give you a side s then the area is s^2. If they give you a diagonal d, then s=d/sqrt(2) so the area is d^2/2.

Oh, okay. Thank you! :D

1. What is the formula for finding the volume of a circle with square cross-sections?

The formula for finding the volume of a circle with square cross-sections is πr^2h, where r is the radius of the circle and h is the height of the cylinder.

2. How do I determine the height of the cylinder if it is not given?

If the height of the cylinder is not given, you can use the Pythagorean theorem to find it. The height will be the square root of the difference between the radius squared and the length of one side of the square cross-section squared.

3. What units should I use for the radius and height when using the formula?

The radius and height should be in the same units. It is important to use consistent units to ensure accurate results.

4. Can I use this formula for any shape of cross-section?

No, this formula is specifically for a circle with square cross-sections. If the cross-section is a different shape, a different formula will need to be used.

5. What could I be doing wrong if my calculated volume seems incorrect?

Double check your calculations and make sure you are using the correct formula. Also, check that your units are consistent and that you have accurately measured the dimensions of the circle and square cross-sections.

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