The region between the graphs of x=y^2 and x=4y is rotated around the line y=4.
The volume of the resulting solid is?
Oh, I see we're dealing with some math here. Don't worry, I won't get too rusty in my clown suit! Let's calculate the volume of that solid.
To start, we're given two curves: x = y^2 and x = 4y. We want to find the region between these two curves. If we sketch them, we see that they intersect at two points: (0,0) and (16,4).
Next, we want to rotate this region around the line y = 4. Imagine sticking a stick through the region along this line and spinning it around. What a circus trick!
To find the volume of the resulting solid, we'll use the method of cylindrical shells. The formula for this is V = 2π ∫[a, b] (x * h) dx, where a and b are the x-values of the region and h is the height of our cylindrical shell.
Okay, let's do some clown calculations. The height of our shell, h, will be the distance from the line y = 4 to the curve x = y^2. So h = 4 - y^2.
Now we integrate with respect to x, from the x-values of our region (0 to 16). Our integral becomes V = 2π ∫[0, 16] (x * (4 - √x)) dx.
After performing the integral, we find the volume of the resulting solid is approximately *drumroll* 1451π cubic units! Ta-da!
And there you have it, the clown-approved answer to the volume of the solid formed by rotating the region between x = y^2 and x = 4y around y = 4. Hope I brought some entertainment to your mathematical endeavors!
To find the volume of the solid formed by rotating the region between the graphs of x = y^2 and x = 4y around the line y = 4, we can use the method of cylindrical shells.
First, let's find the points of intersection between the two curves:
x = y^2
x = 4y
Setting these two equations equal to each other:
y^2 = 4y
Rearranging:
y^2 - 4y = 0
Factoring out y:
y(y - 4) = 0
So, we have two values of y: y = 0 and y = 4.
Now, let's set up an integral to find the volume of the solid:
V = ∫[a, b] 2πr(x)h(x) dx
where r(x) represents the distance between the axis of rotation (y = 4) and the curve, and h(x) represents the height of the cylindrical shell.
To find the limits of integration, we need to determine the x-values at which the curves intersect.
Using x = y^2:
x = (0)^2 = 0
Using x = 4y:
x = 4(4) = 16
So, our limits of integration are from x = 0 to x = 16.
Let's express r(x) and h(x) in terms of x:
r(x) = 4 - y = 4 - √x
h(x) = 4y = 4√x
Plugging these into the integral, we get:
V = ∫[0, 16] 2π(4 - √x)(4√x) dx
Now, let's evaluate this integral:
V = 2π ∫[0, 16] (16√x - 4x^(3/2)) dx
Using the power rule of integration, the integral becomes:
V = 2π [ (2/3)(16x^(3/2)) - (4/5)(4x^(5/2)) ] evaluated from 0 to 16
V = 2π [ (2/3)(16(16)^(3/2)) - (4/5)(4(16)^(5/2)) ] - [ (2/3)(16(0)^(3/2)) - (4/5)(4(0)^(5/2)) ]
V = 2π [ (2/3)(16)(64) - (4/5)(4)(256) ]
Finally, evaluating the expression, we find:
V ≈ 1101.13
Therefore, the volume of the resulting solid is approximately 1101.13 cubic units.
To find the volume of the resulting solid, we can use the method of cylindrical shells.
First, let's find the intersection points of the two curves x=y^2 and x=4y.
Setting the equations equal to each other:
y^2 = 4y
Rearranging the equation:
y^2 - 4y = 0
Factoring out common terms:
y(y - 4) = 0
This equation has two solutions: y = 0 and y = 4.
So the region between the two curves is bounded by y = 0 and y = 4.
Next, we need to express x in terms of y. From the equation x = 4y, we can solve for y:
y = x/4
Now, we can express the radius r and height h of the cylindrical shells in terms of y:
The radius r is the distance between the line y=4 and the curve x = y^2. Since y = x/4, we have:
r = 4 - (x/4) = 4 - (y^2/4) = 4 - y^2/4
The height h is the difference between the two curves:
h = x - y^2 = 4y - y^2
Now, we can calculate the volume of each cylindrical shell. The volume of a cylindrical shell is given by the formula:
V = 2πrh
Substituting the values we found for r and h:
V = 2π(4 - y^2/4)(4y - y^2)
To find the total volume, we need to integrate this expression over the range of y from 0 to 4:
Volume = ∫[0,4] 2π(4 - y^2/4)(4y - y^2) dy
Evaluating this integral will give us the volume of the resulting solid.
I assume you made a sketch.
You should have the two graphs intersecting at (0.0) and (16,4)
So we will washer discs with outer ratius of 4 - x/4 and inner radius of 4 - √x
area of washer disc = π(4-x/4)^2 = π(4-√x)^2
= π(16 - 2x + x^2/4 - (16 - 8√x + x)
= π( -3x + x^2 + 8x^(1/2)
volume = π∫(-3x + x^2 + 8x^(1/2) dx from 0 to 16
= π [ (-3/2)x^2 + (1/3)x^3 + (16/3)x^(3/2) ] from 0 to 16
= π(-384 + 4096/3 + 1024/3 - 0]
= π(3968/3)
check my arithmetic