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The base of S is a circular disk with radius r, Parallel crosssections perpendicular 
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#1
Oct508, 02:01 PM

P: 37

1. The problem statement, all variables and given/known data
The base of S is a circular disk with radius r, Parallel crosssections perpendicular to the base are isosceles triangles with height h, and unequal side in the base. 2. Relevant equations A = 1/2 bh V = integral A(x) from [r, r] 3. The attempt at a solution x^2 +y^2 = r^2 A = 1/2 bh b = 2y = 2sqrt[r^2  x^2] so that gives A = 1/2 [ 2sqrt(r^2  x^2)]h which gives A = sqrt[r^2  x^2]h but i DONT KNOW WHAT h IS I hope I am going in the right directions thanks so much for helping 


#2
Oct508, 03:19 PM

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You don't know what r is either. That shouldn't stop you from integrating dx to find the volume in terms of h and r.



#3
Oct508, 03:28 PM

P: 37

so h is just treated as a constant and the answer equals h* integral of sqrt[r^2x^2]?
my final answer = pi (r^2)h can someone please check my work, thanks ! 


#4
Oct508, 03:44 PM

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The base of S is a circular disk with radius r, Parallel crosssections perpendicular
pi*r^2*h is the volume of a cylinder of height h. The volume should come out to be smaller than that, shouldn't it?



#5
Oct508, 03:50 PM

P: 37

i realized that as well, but when i integrated sqrt[r^2  x^2] from [r,r] and then multiplied that h/2, thats the answer i got. i integrated using the table of integrals so the integration shouldn't be wrong, but u are right, the answer does seem strange
is the this the right way to go about it tho, I can always double check my math, as long as the method is correct 


#6
Oct508, 03:57 PM

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The method is completely correct. How did you get pi*r^2 out of your table? The integral of sqrt(r^2x^2) should give you the area only of the upper half of the circle.



#7
Oct508, 04:11 PM

P: 37

i got the integral of sqrt(r^2x^2) out of the table and in the form of sqrt(a^2u^2)
so the integral = x/2sqrt(r^2x^2) + r^2/2sin1(x/r) from [r,r] plugging r, r into the equation gives you pi from sin^1(r/r) = sin^1(1) = pi/2 and r gives you 3pi/2 the inverse sin is where I got pi from I did forget to mutiply by 2, because you said this is only the upper half of the circle, but is the rest of the integration right? 


#8
Oct508, 04:20 PM

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3pi/2?? You have (r^2/2)*(arcsin(1)arcsin(1)). What's that?



#9
Oct508, 04:27 PM

P: 37

i know, 3pi/2 was just from the arcsin(1)
that woel thing gives you r^2 pi doesn't it (r^2/2) * (pi/2  pi/2) = r^2pi/2 and then multiplied by h and multiplied by 2 to double the volume right? pi*r^2*h 


#10
Oct508, 04:33 PM

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Last time I checked arcsin(1)=pi/2. So yes, the integral sqrt(r^2x^2) is r^2pi/2. Yes, then multiply by h. Then you're done. What other '2' is there to multiply by?



#11
Oct508, 04:44 PM

P: 37

i thought you said that was half the volume so i need to multiply by another 2?



#12
Oct508, 04:49 PM

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No, no, no. I said I knew the integral of sqrt(r^2x^2) wasn't pi*r^2 because it's the area of half a circle. Remember that '2' you cancelled back in the setup with the '1/2' from the triangle area. You did that correctly, the setup was correct. If you were putting arcsin(1)=3/2, that was your problem.



#13
Oct508, 04:56 PM

P: 37

ohhh icic, thank you so much!!!
i just dont know when to use pi/2 or 3pi/2 because they are technically the same 


#14
Oct508, 05:00 PM

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arcsin(x)=y is not just any old y that satisfies sin(y)=x. If it were, it would be hard to evaluate integrals with it. arcsin is specifically DEFINED to have a range of [pi/2,pi/2]. Period.



#15
Oct508, 05:15 PM

P: 37

thank you so much for helping



#16
Oct1208, 11:13 PM

P: 6

i have a problem that is almost identical to bluebear19's so i thought it would be appropriate to post it on this thread, since, Dick, you seem to know it very well
so there is a circle described by the equation x^2+y^2=1 with right isosceles triangles perpendicular to it. each of the triangles is touching the circle with one of its LEGS, with the 90 degree angle and one of the smaller angles touching the perimeter of the circle. now use an integral to find the volume of the solid produced by the adding all of the areas of the infinitely thin triangles from y = 1 to 1. (Just in case my interpretation wasn't as clear to you as it is to me, the actual question is: "The base of the solid is the disk x^2+y^2 <= 1. The crosssections by planes perpendicular to the y axis between y= 1 and y=1 are isosceles right triangles with one LEG on the disk") so i got, V(x)=int from a to b(A(x)dx) A(x)=int from 1 to 1(height of the triangle*base of the triangle) base = x and this is where i get tripped up, if the base, being one of the LEGS of the triangle is x, shouldnt the height be x too? that would make the answer: V(x)=int from 1 to 1((.5x^2)dx) but that would mean that , using .5[(x^3)/3] as the antiderivitive and evaluating from 1 to 1, the answer is 1/3 but if you think about it in a different way , its a half of a can that has a r of 1 and a h of 2 and using (pi*r^2*h)/2 you would get 3.14 which is pi where am i going wrong with this? 


#17
Oct1208, 11:38 PM

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Sure if the base is x, the height is x. But the base ISN'T x, is it? Look at x=0 where the base should be 1. The base is the length of the vertical chord to the circle at x. What is it? BTW, it's probably not a great idea to post questions onto an existing post. It increases the chance nobody will even notice your question.



#18
Oct1308, 12:10 AM

P: 6

yeah i realized that and made a new thread, but so what your saying is
that the base isnt x, using x^2+y^2=r^2 and the fact that using x^2+y^2=1 is the area equation for the base that the base is sqrt(2r^2+x^2)? is that the height as well? 


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