Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis.
y=9-x^2,y=0,x=2,x=3
To find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis, we first need to sketch the region and determine the limits of integration.
The region bounded by the graphs of the equations is a region between the curves y = 9 - x^2 and y = 0, bounded by the vertical lines x = 2 and x = 3.
The limiting values of y are 0 and 9-x^2. The limits of integration for this region are x = 2 and x = 3.
The volume of the solid generated by revolving this region about the y-axis can be calculated using the formula for the volume of a solid of revolution:
V = π∫[a,b] (f(x))^2 dx
where f(x) is the distance of the function from the axis of revolution and the limits of integration are from a to b.
In this case, f(x) = 9 - x^2 and the limits of integration are from x = 2 to x = 3.
So, the volume of the solid is:
V = π∫[2,3] (9 - x^2)^2 dx
V = π∫[2,3] (81 - 18x^2 + x^4) dx
V = π(81x - 6x^3 + (x^5)/5) ∣ [2,3]
V = π[(81(3) - 6(3)^3 + (3)^5/5) - (81(2) - 6(2)^3 + (2)^5/5)]
V = π[243 - 162 + 243/5 - 162 + 48 - 32/5]
V = π[81 + 243/5 + 48 - 32/5]
V = π(336/5 + 48 - 32/5)
V = π(336/5 + 240/5 - 32/5)
V = π(544/5)
V = 1088π/5
Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis is 1088π/5 cubic units.