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

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The discussion focuses on calculating the volume of a solid formed between the curves y = x^2 - 2 and y = 4, with square cross-sections perpendicular to the x-axis. It suggests that Simpson's Rule is unnecessary for this problem, advocating instead for a cross-sectional area approach. The side length of the square is determined by the difference between the two curves, specifically 4 - (x^2 - 2). The volume can be computed using the integral of the area function, leading to the formula ∫ from -2 to 2 of (4 - (x^2 - 2))^2 dx. The final volume calculation results in 16.
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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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...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...
 
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 4-[(x^2)-2]. 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

\int^a_b A(x)\delta x

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

you get the \int^2_{-2} {4-[x^2-2)]}^2 \delta x=16
 
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Thread 'Variable mass system : water sprayed into a moving container'

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