Volume of Solids with Known Cross Sections (Calculus AB)

In summary, the conversation revolves around a calculus problem where the volume of a solid is to be found by revolving the area bounded by two equations around the x-axis and y = 3. The first integral is correct but the second one has some errors in the coefficients. The limits of integration for both integrals are -3 and 0. The conversation also touches upon differentiating between R(x)^2 - r(x)^2 as (top equation minus line)^2 - (bottom equation - line)^2 and (top minus bottom)^2 - (line)^2, with R being the larger radius and r being the smaller radius.
  • #1
Flaneuse
23
0
Can anyone help me with this calculus problem? I have tried it a couple of different ways, but keep getting various incorrect answers; I think my problem is probably in the setup.

I am trying to find the volume of the solid created by revolving the area bounded by these two equations
y = 6 - 2x - x[tex]^{2}[/tex]
y = x + 6
around the x-axis, and around y = 3.

First, I found the intersection points of the two equations, which are at (-3,3) and (0,6)
For the revolution around the x-axis, I tried doing
[tex]\pi \int ^{0}_{-3}((6 - 2x - x[^2}[/tex])[tex]^{2}[/tex] - (x+6)[tex]^{2}[/tex] )dx,

simplified to [tex]\pi [/tex] \int ^{6}_{0}(-32x -13x[tex]^{2}[/tex] +4x[tex]^{3}[/tex] + x[tex]^{4}[/tex] )dx ,
and got a final answer of -27[tex]\pi[/tex]/5
The correct answer is 243[tex]\pi[/tex] /5

For the revolution around y = 3, I got -3-x^{2} from subtracting the lower function, x + 6, from the top function, 6 - 2x - x^{2}, and then tried
[tex]\pi \int^{6}_{0}(6-(-3-x^{2}))^{2} dx, simplified to \pi^{6}_{0}\int (6+3x+x^{2} ) dx, and ended up with 29376[tex]\pi[/tex]/5.

The answer for this one is supposed to be 108[tex]\pi[/tex]/5


Again, I think my real problems are in the setup, probably with the "top minus bottom" part.



Also, I am new to Physics Forum, and can't seem to get rid of how, in the second integral, my words ended up included in the integral. I would appreciate it if anyone could tell me how to fix/avoid that.

Thanks!
 
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  • #2
Note: Fixed up some of your LaTeX by removing a lot of superfluous [ tex] tags.
Flaneuse said:
Can anyone help me with this calculus problem? I have tried it a couple of different ways, but keep getting various incorrect answers; I think my problem is probably in the setup.

I am trying to find the volume of the solid created by revolving the area bounded by these two equations
y = 6 - 2x - x[tex]^{2}[/tex]
y = x + 6
around the x-axis, and around y = 3.

First, I found the intersection points of the two equations, which are at (-3,3) and (0,6)
For the revolution around the x-axis, I tried doing
[tex]\pi \int ^{0}_{-3}((6 - 2x - x^2)^2 - (x+6)^2)dx [/tex]

simplified to [tex]\pi \int ^{6}_{0}(-32x -13x^{2} +4x^{3} + x^{4} )dx [/tex]
The first integral above was fine, but there are some mistakes in the second one. You switched the limits of integration. They should still be -3 and 0. Also, the coefficients of the x^4 and x^3 terms are right, but those for the x^2 and x terms are incorrect.
Flaneuse said:
and got a final answer of -27[tex]\pi[/tex]/5
The correct answer is 243[tex]\pi[/tex] /5

For the revolution around y = 3, I got -3-x^{2} from subtracting the lower function, x + 6, from the top function, 6 - 2x - x^{2}, and then tried
[tex]\pi \int^{6}_{0}(6-(-3-x^{2}))^{2} dx[/tex]
simplified to
[tex]\pi^{6}_{0}\int (6+3x+x^{2} ) dx[/tex]
and ended up with 29376[tex]\pi[/tex]/5.

The answer for this one is supposed to be 108[tex]\pi[/tex]/5


Again, I think my real problems are in the setup, probably with the "top minus bottom" part.



Also, I am new to Physics Forum, and can't seem to get rid of how, in the second integral, my words ended up included in the integral. I would appreciate it if anyone could tell me how to fix/avoid that.

Thanks!
I'll take a look at the 2nd integral in a separate post.
 
  • #3
For the second integral,
[tex]\Delta V = \pi (R^2 - r^2)\Delta x[/tex]
The limits of integration are the same, namely x = -3 to x = 0. R = the y-value on the parabola - 3, and r = the y-value on the line - 3.
 
  • #4
Hm, I don't know what I must have been thinking when I decided to start integrating from 0 to 6, especially when I had already started using -3 to 0 as well... I suppose I was particularly jumbled yesterday, or something of that sort.

Thanks so much for the help; goodness knows I needed it. May I ask though, how can I differentiate between [yes, a very bad choice of words; I mean when to use one as opposed to the other, not take the derivative of] when R(x)^2 - r(x)^2 is equal to (top equation minus line)^2 - (bottom equation - line)^2 and when it is equal to (top minus bottom)^2 - (line)^2?
 
  • #5
Flaneuse said:
Thanks so much for the help; goodness knows I needed it. May I ask though, how can I differentiate between [yes, a very bad choice of words; I mean when to use one as opposed to the other, not take the derivative of] when R(x)^2 - r(x)^2 is equal to (top equation minus line)^2 - (bottom equation - line)^2 and when it is equal to (top minus bottom)^2 - (line)^2?
R is the larger radius and r is the smaller radius. Both radii are as measured from the axis of rotation. If you have drawn a sketch, it should be pretty obvious.
 
  • #6
Thanks!
 

What is the formula for finding the volume of a solid with known cross sections?

The formula for finding the volume of a solid with known cross sections is V = ∫A(x)dx, where A(x) represents the area of the cross section at a given value of x and the integral is taken over the bounds of the solid.

What are some common cross sections used in finding the volume of solids?

Some common cross sections used in finding the volume of solids include squares, rectangles, circles, triangles, and semicircles. These shapes are often used because their areas can be easily calculated using known formulas.

Can the volume of a solid with known cross sections be found using only one integral?

Yes, the volume of a solid with known cross sections can be found using only one integral if the cross sections are perpendicular to the axis of rotation. In this case, the integral can be taken with respect to the variable of integration along the axis of rotation.

What is the difference between a solid with known cross sections and a solid of revolution?

A solid with known cross sections is a three-dimensional shape that can be created by taking cross sections of a two-dimensional shape and stacking them together. A solid of revolution, on the other hand, is created by rotating a two-dimensional shape around an axis to form a three-dimensional shape.

How is the volume of a solid with known cross sections calculated using calculus?

The volume of a solid with known cross sections is calculated by taking the integral of the area of each cross section with respect to the variable of integration. This integral can be evaluated using the Fundamental Theorem of Calculus or by using known formulas for the areas of common cross sections.

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