- #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.
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.
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.
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.
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.
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.