Find the volume of the solid generated by revolving the given region in the 1st quadrant about the y-axis:
y=10x^2,x=0,y=810.
To find the volume of the solid generated by revolving the given region in the 1st quadrant about the y-axis, we can use the method of cylindrical shells.
First, let's understand the region that needs to be revolved. The region is bound by the curve y = 10x^2, the x-axis, the vertical line x = 0, and the horizontal line y = 810. We need to find the volume of the solid formed when this region is rotated about the y-axis.
Step 1: Determine the range of y-values for the region.
The given region is bound by y = 10x^2 and y = 810. Setting these two equations equal to each other, we can find the x-value at which they intersect:
10x^2 = 810
x^2 = 81
x = +/- 9
Since we are only considering the region in the 1st quadrant, we take x = 9 as the upper limit of integration.
Step 2: Setup the integral for the volume using cylindrical shells.
The volume of the solid generated by revolving the region can be calculated by integrating the product of the circumference of each cylindrical shell and its height over the range of y-values.
Since we are revolving about the y-axis, the radius of each cylindrical shell is the x-value of the function y = 10x^2.
The height of each shell is given by dy, which represents an infinitesimally small width of the shell.
Thus, the integral setup is:
V = ∫[a,b] 2π * x * h * dy
where a = 0 and b = 810, x = sqrt(y/10), and h is the width of each cylindrical shell.
Step 3: Evaluate the integral.
V = ∫[0,810] 2π * sqrt(y/10) * h * dy
Since the height of each shell is infinitesimally small, we can replace h with dy.
V = ∫[0,810] 2π * sqrt(y/10) * dy
Integrating with respect to y, we have:
V = 2π * ∫[0,810] sqrt(y/10) * dy
Evaluating this integral will give us the volume of the solid generated by revolving the given region about the y-axis.