The region between the graphs of x=y^2 and x=6y is rotated around the line y=6. Find the volume of the resulting solid please...
i am unable to get the correct answer :(
To find the volume of the solid formed by rotating the region between the graphs of x = y^2 and x = 6y around the line y = 6, you can use the method of cylindrical shells.
Step 1: Draw the graph of the given equations.
The first equation x = y^2 represents a parabolic curve opening towards the right, and the second equation x = 6y represents a linear equation inclined at 45 degrees. Both equations intersect at (36, 6).
Step 2: Determine the boundaries of integration.
To find the limits of integration, we need to identify the points where the two curves intersect.
Setting y^2 = 6y, we can rewrite the equation as y^2 - 6y = 0, which can be factored as y(y - 6) = 0. So, y = 0 or y = 6.
Thus, the range of y will be from 0 to 6.
Step 3: Set up the integral.
The volume of the solid can be found using the formula:
V = ∫(2πrh)dy
In this case, the radius (r) of the cylindrical shell is the distance from the axis of rotation (y = 6) to the curve x = y^2, which is 6 - y. The height (h) is the difference between the x-values of the two curves at a given y-value, which is 6y - y^2.
Now, we can set up the integral:
V = ∫(0 to 6) [2π(6 - y)(6y - y^2)] dy
Step 4: Evaluate the integral.
Evaluating the integral will give us the volume:
V = 2π ∫(0 to 6) [(36y - 6y^2) - (y^3 - 6y^2)] dy
V = 2π ∫(0 to 6) (36y - 6y^2 - y^3 + 6y^2) dy
V = 2π ∫(0 to 6) (-y^3 + 36y) dy
V = 2π [-∫(0 to 6) (y^3 - 36y) dy]
V = 2π [-1/4y^4 + 18y^2] (0 to 6)
V = 2π [-1/4(6^4) + 18(6^2) - (-1/4(0^4) + 18(0^2))]
V = 2π [-1296 + 648]
V = 2π (−648)
V = -1296π
Step 5: Interpret the result.
The volume of the resulting solid is -1296π cubic units.
It is important to note that the negative sign indicates that the solid is oriented below the y-axis. However, in most cases, volume is considered non-negative, so the absolute value of the result, 1296π cubic units, can be used instead.
To find the volume of the resulting solid, we can use the method of cylindrical shells.
Since we are rotating the region between the graphs of x = y^2 and x = 6y around the line y = 6, the vertical axis of rotation will be the y-axis.
To start, let's find the points where the two graphs intersect:
x = y^2
x = 6y
Setting them equal to each other:
y^2 = 6y
Rearranging the equation:
y^2 - 6y = 0
Factoring out y:
y(y - 6) = 0
This equation gives us 2 solutions:
y = 0
y = 6
Now, we need to find the limits of integration for our cylindrical shells. We'll integrate from y = 0 to y = 6.
To find the radius of each shell, we need to find the distance between the line y = 6 and the curve x = y^2.
The equation of the line y = 6 can also be written as x = 0y + 6.
The distance between the line and the curve at a given y-value is given by:
radius = x - (0y + 6)
radius = x - 6
Now, let's write the equation for the volume of a cylindrical shell:
V = 2π∫[from 0 to 6](radius * height) dy
The height of each shell is given by the difference in x-values between the two curves for a specific y-value. So, height = x (curve 1) - x (curve 2).
Plugging in our expressions:
V = 2π∫[from 0 to 6]((x - 6) * (y^2 - 6y)) dy
Now, we need to express x in terms of y. From the equation x = y^2, we get x = y^2.
Substituting this expression into our integral equation:
V = 2π∫[from 0 to 6]((y^2 - 6) * (y^2 - 6y)) dy
Now, we can integrate this expression. The volume will be the result of this integral calculation.