sketch the region R bounded by the graphs of the equations, and find the volume of the solid generated if R is revolved about the x axis when y=x^2 y=4-x^2
The curves intersect where x = ±√2.
By symmetry, using discs,
v = 2∫[0,√2] π(R^2-r^2) dx
where R = 4-x^2, and r = x^2
v = 2π∫[0,√2] ((4-x^2)^2 - (x^2)^2) dx
= 2π∫[0,√2] (16 - 8x^2 + x^4 - x^4) dx
= 2π∫[0,√2] (16 - 8x^2) dx
= 2π(16x - 8/3 x^3)[0,√2]
= 2π(16√2 - 16/3 √2)
= 64π√2/3
To sketch the region R bounded by the graphs of the equations y = x^2 and y = 4 - x^2, we need to find the intersection points between these two equations.
Step 1: Find the intersection points
Set x^2 = 4 - x^2 and solve for x:
2x^2 = 4
x^2 = 2
x = ±√2
So the intersection points are (√2, 2) and (-√2, 2).
Step 2: Sketch the region
Plot the points (√2, 2) and (-√2, 2) on the coordinate plane. These points will be the highest points of each curve. Then, sketch the curves y = x^2 and y = 4 - x^2 symmetrically about the x-axis, passing through these points. The region R will be the area between these curves.
Step 3: Determine the volume of the solid of revolution
To find the volume of the solid generated by revolving R about the x-axis, we can use the method of cylindrical shells. Slice the region R into thin cylindrical shells with thickness Δx. The height of each cylindrical shell will be the difference between the top curve (y = 4 - x^2) and the bottom curve (y = x^2), which is (4 - x^2) - x^2 = 4 - 2x^2.
The volume of each cylindrical shell is given by Vshell = 2πrhΔx, where r is the distance from the x-axis to the shell (in this case, r = x) and h is the height of the shell.
Integrate the volume of each shell from x = -√2 to x = √2:
V = ∫[from -√2 to √2] 2πx(4 - 2x^2) dx
Simplifying the equation and evaluating the integral will give you the volume of the solid generated by revolving R about the x-axis.