Volume by parallel cross-section.

In summary, the base of a solid is bounded by the lines y=x^2 and y=4, and the volume is found by taking the integral of A(x) = (4-x^2)^2 from x=-2 to x=2. The volume element is represented by (4 - x2)(4 - x2)\Delta x, with the first factor being the distance between the two lines, the second factor being the vertical height, and the \Delta x factor being the thickness. The final answer is 512/15 and any calculation errors can be corrected by using the shell method.
  • #1
Call my name
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The questions: the base of a solid is the region bounded y=x^2 and y=4. Find the volume of the solid given that the cross sections perpendicular to the x-axis are squares.

The answer is 512/15.

I set A(x)= (2y)^2 = 4(x^2)^2=4x^4, and the answer is wrong when I integrate this from x = -2 to x = 2.

How can I solve this?

Please help me.
 
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  • #2
The typical volume element, [itex]\Delta V[/itex], has a volume of (4 - x2)(4 - x2)[itex]\Delta x[/itex]. The first factor of (4 - x2) is the distance between the line y = 4 and the curve y = x2. The second factor of (4 - x2) is the vertical height of the volume element (perpendicular to the x-y plane). The [itex]\Delta x[/itex] factor is the thickness of the volume element.
 
  • #3
That's more confusing... can you explain it with integrals?
 
  • #4
Maybe this picture will help you:

forumslab.jpg
 
  • #5
LCKurtz said:
Maybe this picture will help you:

forumslab.jpg

Thank you for the picture, but I do understand how it looks...

just don't know how it goes with the integrals...
 
  • #6
Mark44 said:
The typical volume element, [itex]\Delta V[/itex], has a volume of (4 - x2)(4 - x2)[itex]\Delta x[/itex]. The first factor of (4 - x2) is the distance between the line y = 4 and the curve y = x2. The second factor of (4 - x2) is the vertical height of the volume element (perpendicular to the x-y plane). The [itex]\Delta x[/itex] factor is the thickness of the volume element.

so A(x) = (4-x2)2 ?
2[tex]\int[/tex](4-x2)2dx (x is 0~2)
 
  • #7
Yes, since you have symmetry about x = 0.
 
  • #8
LCKurtz said:
Yes, since you have symmetry about x = 0.

But, I am not getting 512/15.
 
  • #9
Then you have done something wrong.
[tex]\int_{x = -2}^2 (4 - x^2)^2 dx~=~2\int_{x = 0}^2 (4 - x^2)^2 dx~=~\frac{512}{15}[/tex]
 
  • #10
Yes. made some calculation error
 
  • #12
Mark44 said:
Did you find out where you went wrong?

Yes. Thank you. I have posted another question this time regarding shell method
 

Related to Volume by parallel cross-section.

1. What is volume by parallel cross-section?

Volume by parallel cross-section is a method of calculating the volume of an object by taking cross-sectional slices of the object and adding them together. This is often used for irregularly shaped objects or objects with varying cross-sectional areas.

2. How is volume by parallel cross-section different from other methods of calculating volume?

Unlike other methods, volume by parallel cross-section takes into account the varying cross-sectional areas of an object, making it more accurate for irregularly shaped objects.

3. What are some real-world applications of volume by parallel cross-section?

Volume by parallel cross-section is commonly used in fields such as architecture, engineering, and manufacturing to calculate the volume of irregularly shaped objects or structures.

4. How do you calculate volume by parallel cross-section?

To calculate volume by parallel cross-section, you first need to take measurements of the cross-sectional areas of the object at regular intervals. Then, you add these areas together and multiply by the distance between each cross-section to get the total volume.

5. What are the limitations of volume by parallel cross-section?

Volume by parallel cross-section may not be accurate for objects with complex internal structures or irregularly spaced cross-sections. Additionally, it may be difficult to obtain precise measurements for the cross-sectional areas, leading to potential errors in the calculation.

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