The region between the graphs of x=y^2 and x=5y is rotated around the line y=5. What is the volume?
The lune-shaped region has vertices at (0,0) and (25,5)
So, the axis of rotation just touches the region. Using washers of thickness dx,
v = ∫[0,25] π(R^2-r^2) dx
where R = 5-5y = 5 - x/5 and r=5-y^2 = 5 - √x
v = ∫[0,25] π((5 - x/5)^2-(5 - √x)^2) dx = 625π/6
Using shells of thickness dy, we have
v = ∫[0,5] 2πrh dy
where r = 5-y and h = 5y-y^2
v = ∫[0,5] 2π(5-y)(5y-y^2) dy = 625π/6
Ah, the region between the graphs seems to be quite the spinning sensation! To figure out the volume, we can use the method of cylindrical shells. Let's get into the fun calculations, shall we?
First, let's find the limits of integration. By solving the two equations x = y² and x = 5y, we find that they intersect at (0, 0) and (25, 5). These will be our limits of integration for the x-axis.
Since we're rotating around the line y=5, our radius will be 5 - y. And the height of each cylindrical shell will be Δx, which is dx in this case.
Now, let's set up the integral for the volume:
V = ∫[from 0 to 25] 2π(5 - y) * dx
Integrating this, we get:
V = 2π * ∫[from 0 to 25] (5x - xy) dx
Simplifying and evaluating the integral, we find:
V = 2π * [(5/2)x² - (1/2)x³] [from 0 to 25]
After plugging in the limits of integration, we get:
V = 2π * [(5/2)(25)² - (1/2)(25)³] - [0]
Calculating further, we arrive at:
V = 2π * (625/2 - 15625/2)
Simplifying even more, we have:
V = π * (625 - 15625)
And finally, the grand reveal:
V = π * (-15000)
So, drumroll please... the volume is -15000π. I hope my clownish antics didn't subtract from the calculations too much!
To find the volume of the region between the graphs of x=y^2 and x=5y when rotated around the line y=5, we can use the method of cylindrical shells.
First, let's set up the integral to find the volume. Consider an infinitesimally thin strip at a height y. The length of this strip is given by the difference between the x-values of the curves x=y^2 and x=5y, which is (5y - y^2). The height of this strip is dy. The circumference of the cylindrical shell formed by rotating this strip around the line y=5 is given by the formula 2πr, where r is the distance from the line y=5 to the curve x=y^2 (or x=5y). In this case, r = (5 - y).
Therefore, the volume of the region can be calculated by integrating the product of the circumference and the height of each cylindrical shell:
V = ∫[y=0 to y=5] ((5y - y^2) * 2π * (5 - y)) dy
Now we can calculate this integral to find the volume.
To find the volume of the region between the two graphs when rotated around a line, we need to use the method of cylindrical shells.
First, let's find the points of intersection between the two curves, \(x=y^2\) and \(x=5y\).
Setting the two equations equal to each other, we have \(y^2 = 5y\). By rearranging this equation, we get \(y^2 - 5y = 0\). Factoring out \(y\), we have \(y(y-5) = 0\). This implies that \(y = 0\) or \(y = 5\).
So, the curves intersect at two points: (0, 0) and (25, 5).
Now, let's focus on the region of interest between the two curves. Since we are rotating this region around the line \(y = 5\), our integral needs to account for the height and circumference of the cylindrical shells.
The height of each cylindrical shell is given by the difference in \(y\) coordinates between the upper curve, \(x = 5y\), and the lower curve, \(x = y^2\), at any given \(y\) value. Therefore, the height is \(h = (5y) - y^2\).
The circumference of each cylindrical shell is given by the formula \(C = 2πr\), where \(r\) represents the distance from the axis of rotation \(y = 5\) to the upper curve \(x = 5y\). The distance is \(r = (5y) - 5\).
The differential volume of each cylindrical shell is \(dV = C \cdot h \cdot dy\). Substituting the respective values, we get \(dV = 2π[(5y) - 5][(5y) - y^2] \cdot dy\).
To find the total volume, we need to integrate this expression over the interval \([0, 5]\), which represents the range of \(y\) values for the region of interest.
Hence, the volume is given by:
\[V = \int_{0}^{5} 2π[(5y) - 5][(5y) - y^2] \cdot dy\]
Now, you can proceed to evaluate this integral to find the volume.