If the area between the curves y=√x and y=x^3
is rotated completely about the x-axis,find the
volum of the solid made?
Please show step
infinite as stated
First step: draw the graphs!
They intersect at (0,0) and (1,1). You want to rotate a small lens-shaped area.
To find the volume, you can do two methods.
Think of the volume as a stack of discs with holes (washers). The holes are there because a slice of area is rotated around the x-axis, but does not touch it. So, with discs of thickness dx, the volume is
v = ∫[0,1] π(R^2-r^2) dx
where R=√x and r=x^3
v = ∫[0,1] π(x-x^6) dx = 5π/14
Or, you can think of the volume as a set of nested cylinders (shells) of thickness dy. Their height is the distance between the two curves x=y^2 and x=∛y.
v = ∫[0,1] 2πrh dy
where r=y and h=∛y-y^2
v = ∫[0,1] 2πy(∛y-y^2) dy = 5π/14
To find the volume of the solid made by rotating the area between the curves y = √x and y = x^3 about the x-axis, you can use the method of cylindrical shells. Here are the steps to find the volume:
Step 1: Determine the limits of integration.
To find the limits of integration, set the two curves equal to each other and solve for x:
√x = x^3
Squaring both sides, we get:
x = x^6
Rearranging, we have:
x^6 - x = 0
Factoring out an x, we get:
x(x^5 - 1) = 0
This equation has two solutions: x = 0 and x = 1.
So the limits of integration are 0 and 1.
Step 2: Set up the integral.
The volume of the solid can be found by integrating the product of the circumference of the shell and its height over the interval [0, 1]:
V = ∫(2πr)(h)dx
The radius of each shell is x, and the height is the difference between the two curves, i.e., h = √x - x^3.
Step 3: Evaluate the integral.
The integral becomes:
V = ∫(0 to 1) 2πx(√x - x^3) dx
To evaluate this integral, we need to expand and simplify the expression:
V = 2π ∫(0 to 1) (x^(3/2) - x^(7/2)) dx
Now, we can integrate each term separately:
V = 2π [ (2/5)x^(5/2) - (2/9)x^(9/2) ] evaluated from 0 to 1
Plugging in the upper and lower limits of integration:
V = 2π [ (2/5)(1^(5/2)) - (2/9)(1^(9/2)) ] - 2π [ (2/5)(0^(5/2)) - (2/9)(0^(9/2)) ]
Simplifying further:
V = 2π [ (2/5) - (2/9) ]
V = 2π [ (18/45) - (10/45) ]
V = 2π [ 8/45 ]
V = (16π)/45
So the volume of the solid made by rotating the area between the curves y = √x and y = x^3 about the x-axis is (16π)/45 cubic units.