Find the volume of the solid with a circular base of radius 9 and the cross sections perpendicular to the y-axis are squares.
I've never solved a problem like this before. How can I go about setting up the integral for this particular problem?
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Is it a transition piece? Before integrating can you describe it more and/or sketch it as much as you can? – Narasimham Jul 17 '22 at 23:45
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@Narasimham That's also why I'm a bit confused. The professor has thrown this problem in an assignment the way I've written it above. I don't think it's a transition piece. I wish I could describe/sketch it more, but I'm still trying to figure out how to set up/approach the problem. – Roman Arefov Jul 17 '22 at 23:51
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1This should give an idea of what the solid looks like. – eyeballfrog Jul 17 '22 at 23:59
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So because of symmetry, split the solid in half along a diameter. This simplifies things.
the cross sections perpendicular to the y-axis are squares
We can thus slice this solid into vertical squares. Have you ever chopped or seen someone chopped a vegetable like a cucumber like in the video below? It is similar to that.
https://www.youtube.com/watch?v=EXgchTrIgw4
So the side length of each slice would be a chord of the circle parallel to the y-axis. The area of each slice would be the square of that. To find the volume of the half-solid we would simply add up the areas of the slices, which is simply integration. So we can integrate the half-solid like this:
$\int_{0}^{9} (2*\sqrt{81 - x^{2}})^2 dx$
From the Pythagorean theorem and area of a square
And then we multiply the above integral by 2.
Since this is an assignment I assume you are in a calculus class or know how to do integration for this problem. Good luck!
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1Thank you for taking the time to answer and explain this. I attempted to solve this before you posted your answer and I got a different answer. This is a picture of my attempt at a solution: https://imgur.com/a/FplaDlO. Since the the cross-sections are perpendicular to the y-axis, don't we have to integrate with respect to y? Please take a look at my solution and let me know what you think. – Roman Arefov Jul 18 '22 at 00:47
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1I am very sorry, I tend to make silly mistakes. Yes, your approach is correct, I edited my answer. Thank you! – MeltedStatementRecognizing Jul 18 '22 at 01:08
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1Also because of symmetry I don't think it matters for x-axis vs y-axis. The base is a circle. And also the x in dx is a integration variable, not for x-axis – MeltedStatementRecognizing Jul 18 '22 at 01:09
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Could not quite follow how volume was computed. Should it not be $\int (2*\sqrt{81 - x^{2}})^2 dz?$ – Narasimham Jul 18 '22 at 17:33
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@Narasimham, That is the indefinite integral used. However, to find the volume you have to use the definite integral between two points in order to get a numerical answer. – MeltedStatementRecognizing Jul 18 '22 at 17:47
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An indefinite integral is the antiderivative of an equation, and is another equation. The definite integral is the area under the graph of an equation and is a number. The two are linked by the Fundamental Theorem of Calculus #2 – MeltedStatementRecognizing Jul 18 '22 at 17:50
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@Narasimham were you talking about the dz in your integral? – MeltedStatementRecognizing Jul 18 '22 at 17:58
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Yes, that should be so in general, but am afraid being suggested towards a chat. – Narasimham Jul 18 '22 at 18:01
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There are many possibilities. Take a look at a solid carved out of a unit sphere as a particular case.
When two planes $\dfrac{z}{x}=2$ intersect a sphere radius $\sqrt 5 $ we have two red great circle intersections and all sections normal to y-axis between these intersection great circles are squares.
The squares stand over $xOy$ plane parallel with an edge on this plane being parallel to $xOz$ plane between ellipse projections as shown in a 3d hand sketch below. Since square discs are stacked integration should proceed with resp to $y$ axis, with elemental volumes obtained by integration with $dy$.
Parameterization of the spheres radius $\sqrt 5$ centered at origin:
$$P(x,y,z)= (\pm t ,\sqrt5 \sqrt{1-t^2}, 2t ) $$
$$ y= \sqrt5 \sqrt{1-t^2} ; dy =\frac{\sqrt5 t dt}{\sqrt{1-t^2}} $$
$$ dV= 4 t^2. dy =4 t^2\frac{\sqrt5 t dt}{\sqrt{1-t^2}}$$ $$ V = 4 {\sqrt5 }\int _0^1\; t^3 {\sqrt{1-t^2}} dt; \; $$
Hope you can take it from here ( substitution $t^2=u$) scaling radius from $\sqrt5$ to 9 by cubing this ratio for volume multiplication and then doubling it for the $-y$ part of the solid.
Narasimham
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