Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
x=9y^2 y=1 x=0 about the y-axis.
I am able to find the volume of other questions, but this one got me stumped. I tried drawing out the graph, but I can't really form the boundaries. help? thank you
The solution is easier than I thought. I just did integral from [0,1] pi*(9y^2)^2 dy
did you get 81pi/5 ?
To find the volume of the solid obtained by rotating the region bounded by the curves x = 9y^2, y = 1, and x = 0 about the y-axis, we can use the method of cylindrical shells.
Let's first plot the given curves on a graph. The curve x = 9y^2 is a parabola that opens to the right. The line y = 1 is a horizontal line at y = 1. The line x = 0 is the y-axis.
To visualize the region bounded by these curves, consider the following:
1. The curve x = 9y^2 starts at the origin (0,0) and extends towards positive x-direction as y increases.
2. The line y = 1 is a horizontal line located above the x-axis at y = 1.
3. The line x = 0 is the y-axis.
Based on this information, we can determine the boundaries of the region we need to rotate about the y-axis. The lower boundary is the y-axis (x = 0), and the upper boundary is the parabola x = 9y^2. The vertical strip between y = 0 and y = 1 represents the region we need to calculate the volume for.
Now, let's set up the integral to find the volume using cylindrical shells method.
Consider a thin vertical strip with thickness dy at a distance y from the y-axis. The height h of this strip is given by h = x = 9y^2, and the radius r is simply the distance from the y-axis to the strip, which is r = y.
The volume of this thin shell is given by dV = 2πrh dy, where 2π is the circumference of a circle.
To find the total volume, we integrate the expression for dV between the limits of y = 0 and y = 1:
V = ∫[from y=0 to y=1] 2π(9y^2)(y) dy
Simplifying the expression inside the integral, we get:
V = 18π ∫[from y=0 to y=1] y^3 dy
Integrating this expression, we get:
V = 18π [y^4/4] [from y=0 to y=1]
Plugging in the limits, we get:
V = 18π [(1/4) - (0/4)]
V = 18π/4
V = 9π/2
Therefore, the volume of the solid obtained by rotating the region bounded by the curves x = 9y^2, y = 1, and x = 0 about the y-axis is 9π/2 cubic units.