The region bounded by y=3x, y=0, x=3, and x=5 is rotated about the x-axis. Find the volume of the resulting solid.
However, just so you get to use your calculus to analyze the frustrum, just think of it as a stack of thin discs of thickness dx:
v = ∫πr^2 dx
where r=y=3x
v = ∫[3,5] π(3x)^2 dx = 294π
Or, as a bunch of nested shells, it gets a bit more complicated, since the interior is a solid cylinder of radius 9 and height 2. The volume of the portion bounded by y=3x is
∫[9,15] 2πrh dy
where r=y and h=(5-x)=(5-y/3)
∫[9,15] 2πy(5-y/3) dy = 132π
add that to the cylinder (162π) and you get 294π
no need for calculus.
your solid is a fulcrum formed by the following:
You have a cone on its side, (the x-axis is the axis of the cone)
height = 3, radius = 9
the other :
height = 5, radius = 15
V = (1/3)π(15^2)(5) - (1/3)π(9^2)(3)
= (1/3)π (1125 - 243)
= 882π/3
= 294π cubic units
To find the volume of the resulting solid, we need to use the method of cylindrical shells.
First, let's sketch the region bounded by the given equations:
Since y=3x and y=0, we can find the x-values by setting 3x=0:
x=0
So, the region is bounded by the lines x=0, x=3, x=5, and the x-axis.
Now, let's consider an infinitely thin vertical strip within this region. This strip will have a width of dx and its height will be the difference between the y-values of the two functions y=3x and y=0.
The height of the strip is given by:
height = (3x - 0) = 3x
To find the circumference of the cylindrical shell, we use the formula:
circumference = 2πr = 2πx
Finally, we can find the volume of the solid by integrating the product of the circumference and the height over the interval from x=0 to x=3:
V = ∫[0,3] (2πx) * (3x) dx
Let's integrate this expression to find the volume.
To find the volume of the resulting solid, we can use the method of cylindrical shells.
The region is bounded by the lines y = 3x, y = 0, x = 3, and x = 5.
First, let's sketch the region.
The lines y = 3x, y = 0, x = 3, and x = 5 form a right triangle with base 3 and height 15.
To find the volume, we need to integrate the area of the cylindrical shells that make up the solid. Each shell has a height of dy and a radius of x (which varies with y).
The radius can be found by solving the equation y = 3x for x:
x = y/3.
The height of the cylindrical shell is dy.
The differential volume of a cylindrical shell is given by the formula:
dV = 2πrh dy,
where r is the radius and h is the height.
Substituting the values, we have:
dV = 2π(y/3)(dy).
Now we can set up the integral and find the volume:
V = ∫(from y = 0 to y = 15) 2π(y/3) dy.
Integrating, we have:
V = (2π/3) ∫(from 0 to 15) y dy.
Evaluating the integral, we have:
V = (2π/3) [y^2/2] (from 0 to 15).
V = (2π/3) [(15^2/2) - (0^2/2)].
V = (2π/3) [225/2].
V = (450π/6).
Simplifying, we have:
V = 75π.
Therefore, the volume of the resulting solid is 75π cubic units.