Volume Question with cirular base and square top

ktpr2

The question goes like this:

"The base of a certain solid is circular with a radius of 1. Cross-sections of the solid, perpendicular to the base, are square. Find the volume of the solid."

In visualizing this object, i see that it looks like a "blocky" foot ball, where the perimeter is circular, but the corners are sharp because of the volume of the reducing square outward from the center.

I center the circle at zero on the xy plane. Then I note that base of a given slab can be given by $$2 \sqrt{1-x^2}$$. Since the area of a sqaure is b^2, $$A(x) = 4(1-x^2)$$. The slab has a width $$\Delta x$$, making the total volume function $$4 (1-x^2) \Delta x$$. I then integrate from 0 to 1 and double the resultin volume, as 0 to 1 represents the volume of half this shape. My result is $$2 \int_{0}^{1} 4 (1-x^2) \Delta x = \frac{32}{3}$$ cubic units.

Is this logic correct? If not, why?

Related Introductory Physics Homework Help News on Phys.org

ehild

Homework Helper
ktpr2 said:
The question goes like this:

"The base of a certain solid is circular with a radius of 1. Cross-sections of the solid, perpendicular to the base, are square. Find the volume of the solid."

In visualizing this object, i see that it looks like a "blocky" foot ball, where the perimeter is circular, but the corners are sharp because of the volume of the reducing square outward from the center.
Look at the picture. What is the name of that shape of solid? Last edited:

ramollari

Why can't the imaginary shape be a cylinder with height equal to the diameter of the base?

PS: Ehild already mentioned it.

ktpr2

Um, isn't that just a cylinder then? And I suppose i introduced an extra constraint; the cross sectino is never said to decrease.

ehild

Homework Helper
ktpr2 said:
Um, isn't that just a cylinder then? And I suppose i introduced an extra constraint; the cross sectino is never said to decrease.
It is!!!! :rofl:

ehild

HallsofIvy

Homework Helper
The figure in the original problem is not a cylinder because, if it were, the cross sections would be rectangles, not squares (the cross section at the diameter would be a square but as you move toward the ends of the diameter at right angles to your cross sections, the base length gets smaller- with a cylinder the height doesn't).
ktpr2 was right the first time- it's a sort of "blocky" football.

Set up a coordinate system so that the origin is at the center of the circular base. The equation of the circle is x2+ y2= 1. Imagine the cross-sections to be perpendicular to the x-axis (so parallel to the y-axis). The base of each cross-section runs from the circle above the x-axis to the circle below the x-axis: the length is 2y and so the area of the square cross-section is 4y2. That is: 4(1- x2). The "thickness" of each imaginary cross-section is dx (since we are "moving" along the x-axis). The "sum" of those becomes the integral $$4\int_{-1}^{1}(1- x^2)dx$$

Last edited by a moderator:

OlderDan

Homework Helper
ktpr2 said:
"The base of a certain solid is circular with a radius of 1. Cross-sections of the solid, perpendicular to the base, are square. Find the volume of the solid."

In visualizing this object, i see that it looks like a "blocky" foot ball, where the perimeter is circular, but the corners are sharp because of the volume of the reducing square outward from the center.

I center the circle at zero on the xy plane. Then I note that base of a given slab can be given by $$2 \sqrt{1-x^2}$$. Since the area of a sqaure is b^2, $$A(x) = 4(1-x^2)$$. The slab has a width $$\Delta x$$, making the total volume function $$4 (1-x^2) \Delta x$$. I then integrate from 0 to 1 and double the resultin volume, as 0 to 1 represents the volume of half this shape. My result is $$2 \int_{0}^{1} 4 (1-x^2) \Delta x = \frac{32}{3}$$ cubic units.

Is this logic correct? If not, why?
There is nothing wrong with your logic, but you need to check your integration. This shape is not a cylinder. Also, when writing integrals we use differentials, not deltas

$$V = 2 \int_{0}^{1} 4 (1-x^2) dx = 8 \int_{0}^{1} (1-x^2) dx = \ \ \????$$

OOPs. Halls beat me to it.

Last edited:

ehild

Homework Helper
HallsofIvy said:
The figure in the original problem is not a cylinder

Sorry, I overlooked the "square top". Does that shape then look like in the picture? Now I am not sure, either, what "cross section" is.

ehild

Last edited:

OlderDan

Homework Helper
ehild said:
Sorry, I overlooked the "square top". Does that shape then look like in the picture? Now I am not sure, either, what "cross section" is.

ehild
That's really not it either. It does not have a flat top. It's more like a cylinder of length 2 and radius 1 that has been cut by a saw with a blade in the shape of half a 2 by 4 ellipse with the long axis of the ellipse parallel to the axis of the cylinder. You have to use a little imagination, but the first figure is an approximation showing only the positive x half of the solid. The corners of the squares are rounded in the figure, as if the cut out shape had the sharp edges sanded round. The actual shape has sharp corners. Draw vertical lines from the rounded corners down the the base to complete the square cross sections. Those extended lines to the base would lie on the circle at the base.

I changed the coordinates so that I could do a better looking plot. The second figure is a pretty fair rendition of the shape of half of the solid. The third shows just about all of it.

Attachments

• 9.4 KB Views: 333
• 49.5 KB Views: 377
• 55.9 KB Views: 405
Last edited:

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving