Find the volume of the solid generated by revolving the region bounded by the given curves and line about the y-axis.
y=50-x^2
y=x^2
x=0
as I see it, the upper and lower boundries cross at 5,25, os the integral will be from x=0 t0 5
so the dArea will be [(50-x^2)-x^2]dx
and you will rotate that about the y axis, so the volume will be INT 2PI*xdA or
int 2PI x(50-2x^2)dx
so integrate that. I will be happy to check it.
Like all multiple integral problems, start with drawing a sketch of the bounding curves.
I have done that for you for this time, see:
http://img687.imageshack.us/img687/811/1342739949.png
If curves intersect, find the intersection points. In this case, it is at (5,25).
Then decide how you want to integrate, namely the order of integrating along x first, followed by y, or vice versa.
Using the ring method, and set up the double integral:
Volume
=∫∫2πx dy dx
y goes from x^2 to 50-x^2 and
x goes from 0 to 5 (intersection point).
2Pi(500/3)
To find the volume of the solid generated by revolving the region bounded by the curves and line about the y-axis, we can use the method of cylindrical shells.
Step 1: Draw the region and the axis of rotation to visualize the problem. In this case, the region is bounded by the curves y = 50 - x^2, y = x^2, and the line x = 0.
Step 2: Determine the interval of integration. Since the region is symmetric about the y-axis, we can integrate from y = 0 to y = 50.
Step 3: Express the radius and height of a typical cylindrical shell in terms of y. The radius of a cylindrical shell is the distance from the axis of rotation to the curve. In this case, the radius is given by r = x. Since we are revolving about the y-axis, we need to rewrite the equations in terms of x. So, y = 50 - x^2 becomes x = sqrt(50 - y) and y = x^2 remains the same. The height of a cylindrical shell is the difference between the y-values of the two curves. In this case, the height is given by h = (50 - x^2) - x^2 = 50 - 2x^2.
Step 4: Set up the integral to find the volume. The volume of a cylindrical shell is given by V = 2πrhΔy, where Δy is the width of the shell in the y-direction. In this case, Δy is the infinitesimally small change in y. Therefore, the volume can be calculated by integrating V = 2π(50 - 2x^2)(x)dy over the interval from y = 0 to y = 50.
Step 5: Evaluate the integral. Integrate V = 2π(50 - 2x^2)(x)dy with respect to y from 0 to 50.
The final result will be the volume of the solid generated by revolving the region bounded by the given curves and line about the y-axis.