the region bounded by the graph f(x)=x(2-x) and the x axis is revolved about the y axis. Find the volume of the solid. I did the integral using the shell method, but the answer wasn't correct.
Why don't you show your work?
Using shells, try
v = ∫[0,2] 2πrh dx
where r = x and h=y=2x-x^2
v = ∫[0,2] 2πx(2x-x^2) dx
= 8π/3
Using washers, we have
v = ∫[0,1] π(R^2-r^2) dy
where r = 1-√(1-y^2) and R = 1+√(1-y^2)
v = ∫[0,1] π((1+√(1-y^2))^2-(1-√(1-y^2))^2) dy
which is a little more complicated.
My bad. Using discs, the integral is
v = ∫[0,1] π(R^2-r^2) dy
where r = 1-√(1-y) and R = 1+√(1-y)
v = ∫[0,1] π((1+√(1-y))^2-(1-√(1-y))^2) dy
= ∫[0,1] 4π√(1-y) dy
= 8π/3
i did get 8pi/3 but it wasn't one of the choices. Maybe there is a mistake in the question
To find the volume of the solid generated by revolving the region bounded by the graph of the function f(x) = x(2-x) and the x-axis about the y-axis, we will need to use the method of cylindrical shells correctly. Let's go through the steps together.
First, let's sketch the region to visualize it better. The graph of f(x) = x(2-x) is a downward-opening parabola that intersects the x-axis at x = 0 and x = 2.
To use the method of cylindrical shells, we need to express the function f(x) in terms of y (inverted). By rearranging the equation, we have:
y = x(2-x)
x² - 2x + y = 0
Now, let's solve for x in terms of y:
x = (2 ± √(4 - 4y))/2
x = 1 ± √(1 - y)
Since we are revolving the region about the y-axis, the radius of each shell will be the distance from the y-axis to the curve at a given value of y. This distance is given by the expression x = 1 ± √(1 - y).
Now, we can write the integral to calculate the volume using the cylindrical shell method:
V = ∫[a,b] 2πx · h · dx
Where a and b are the y-values that correspond to the region bounded by the function f(x) = x(2-x) and the x-axis. To find these limits, let's set the function equal to zero:
0 = x(2-x)
x = 0 or x = 2
Therefore, the limits of integration will be a = 0 and b = 2.
Now, let's substitute the formula for the height (h) of each shell into the integral:
V = ∫[0,2] 2π(1 ± √(1 - y)) · y · dx
Since we have already expressed everything in terms of y, we can rewrite the integral as:
V = ∫[0,2] 2π(1 ± √(1 - y)) · y · dy
Now, it's important to double-check if you have set up and solved the integral correctly. Once you evaluate the integral, you should have the correct volume of the solid.
If you're still having trouble getting the correct answer, please let me know the result of your integral, and I can help you identify where the error may have occurred.