Find the volume of the solid generated by revolving the region bounded by the graphs of the equations and/or inequalities about the indicated axis.
x = y(25 − y2)^1/2
x = 0
the y-axis
To find the volume of the solid generated by revolving the region bounded by the given equations about the y-axis, we can use the method of cylindrical shells.
First, let's sketch the region bounded by the given equations. The equation x = y(25 - y^2)^(1/2) represents the curve that forms the boundary of the region. The second equation, x = 0, represents the y-axis, which is a vertical line passing through the origin.
We can rewrite the equation x = y(25 - y^2)^(1/2) as follows:
x^2 = y^2(25 - y^2)
x^2 = 25y^2 - y^4
Therefore, we have a symmetric region bounded by the curves x = y(25 - y^2)^(1/2) and x = 0.
To find the volume using cylindrical shells, we will integrate from y = 0 to y = a, where 'a' is the y-coordinate of the highest point where the curves intersect.
To find the intersection points, we set the two equations equal to each other:
y(25 - y^2)^(1/2) = 0
y = 0
So, the region is bounded by the y-axis (x = 0) and the x-axis (y = 0) with no intersection points.
Since there are no intersection points and the region is bounded by the x and y-axes, the volume of the solid generated by revolving this region about the y-axis is zero.