Volume of Solid Formed by y=x^2-2 & y=4

  • Thread starter bigskilly
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In summary, the conversation is about finding the volume of a solid formed by a base bounded by two given equations, using squares perpendicular to the x-axis. One person suggests using Simpson's Rule, while the other suggests finding the volume using cross-sections and developing an equation for the area of a square. The final formula for volume by cross-sections is given as \int^2_{-2} {4-[x^2-2)]}^2 \delta x=16.
  • #1
bigskilly
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Find the volume of the solid formed with a base bounded by y = (x^2)-2 and y=4 filled with squares that are perpendicualr to the x-axis.
 
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  • #2
...i think you can find that area using simpson's rule , you can thus proceed to find the volume using the maximum height...yeah i think that should work...
 
  • #3
It's really not necessary to use Simpson's Rule. Find an volume by cross-section. In order to do that, you need to develop an equation for the area, in this case a square. By the equations you gave, the length of one side of the square would be [tex]4-[(x^2)-2][/tex]. Since the cross-sections are perpendicular to the x axis. You can leave the function as is since it is already in terms of x. So the formula for volume by cross-sections is

[tex]\int^a_b A(x)\delta x[/tex]

so after finding your a and b, (set the equations equal to each other)

you get the [tex]\int^2_{-2} {4-[x^2-2)]}^2 \delta x=16[/tex]
 
Last edited:

Related to Volume of Solid Formed by y=x^2-2 & y=4

1. What is the equation for the volume of the solid formed by y=x^2-2 and y=4?

The equation for the volume of the solid formed by these two functions is V = ∫[4 - (x^2-2)]dx, where x ranges from the x-intercepts of x^2-2 to the x-intercept of y=4.

2. How do you find the x-intercepts of y=x^2-2 and y=4?

The x-intercepts of y=x^2-2 can be found by setting y=0 and solving for x, which gives x=±√2. The x-intercept of y=4 is simply x=0.

3. What is the domain and range of the functions y=x^2-2 and y=4?

The domain of both functions is all real numbers, since there are no restrictions on the values of x. The range of y=x^2-2 is y≥-2, since the minimum value of the function is at x=0, and the range of y=4 is y=4, since the function is a horizontal line.

4. Can you graph the functions y=x^2-2 and y=4 and show the solid formed?

Yes, the graph of y=x^2-2 is a parabola opening upwards with a vertex at (0,-2). The graph of y=4 is a horizontal line at y=4. The solid formed by these two functions is the region between the parabola and the horizontal line, bounded by the x-axis. See image below.

5. How can the volume of the solid formed by y=x^2-2 and y=4 be calculated?

The volume can be calculated using the equation V = ∫[4 - (x^2-2)]dx or by using the formula for the volume of a solid of revolution, V = ∫π(y)^2dx, where y=x^2-2 and y=4 are the two functions being rotated around the x-axis. The two methods should yield the same result.

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