Find the volume generated when the area bounded by the curve y^2 = 16x, from x = 0 to x = 4 is rotated around the x-axis.
using discs:
v = ∫[0,4] πr^2 dx
where r=y
v = ∫[0,4] πy^2 dx
= π∫[0,4] 16x dx
= 16π∫[0,4] x dx
= 16π(1/2 x^2)[0,4]
= 16π(8-0)
= 128π
using shells:
v = ∫[0,8] 2πrh dy
where r = y and h = 4-x = 4-y^2/16
v = 2π∫[0,8] y(4 - y^2/16) dy
= 2π (2y^2 - y^4/64)[0,8]
= 2π(128 - 64)
= 128π
To find the volume generated when the area bounded by the curve y^2 = 16x, from x = 0 to x = 4 is rotated around the x-axis, we can use the method of cylindrical shells.
First, let's express the equation y^2 = 16x in terms of y:
y = ±√(16x)
We are given that the bounds of x are from 0 to 4.
Next, we need to find the height of each cylindrical shell. This is given by the difference in y-values at a given x:
height = √(16x) - (-√(16x))
= 2√(16x)
= 4√x
Finally, we can use the formula for calculating the volume generated by rotating a curve around the x-axis using cylindrical shells:
V = ∫[0 to 4] of (2πx)(4√x) dx
Now, let's integrate the expression:
V = 2π ∫[0 to 4] of (4√x^3) dx
Simplifying further:
V = 2π ∫[0 to 4] of (4x^(3/2)) dx
Integrating this expression:
V = 2π [(2/5)(x^(5/2))]|_[0 to 4]
Substituting the bounds:
V = 2π [(2/5)(4^(5/2)) - (2/5)(0^(5/2))]
Calculating the value:
V = 2π [(2/5)(32 - 0)]
V = 2π [(64/5)]
V = 128π/5
Therefore, the volume generated when the area bounded by the curve y^2 = 16x, from x = 0 to x = 4 is rotated around the x-axis is (128π/5) cubic units.
To find the volume generated by rotating the area bounded by the curve around the x-axis, we can use the method of cylindrical shells.
First, let's sketch the curve y^2 = 16x.
This equation represents a parabola that opens to the right, with the vertex at the origin (0,0). The parabola intersects the x-axis at x = 0 and x = 16.
Next, we need to find the limits of integration. In this case, since we are rotating around the x-axis, the limits are from x = 0 to x = 4.
Now, let's set up the integral for the volume using the cylindrical shell method.
The volume can be expressed as:
V = ∫(2πrh)dx
where r is the radius of the shell and h is the height of the shell. In this case, the radius is equal to x, and the height is equal to y.
From the equation y^2 = 16x, we can solve for y:
y = ±√(16x) = ±4√x
Given that we are looking for the volume when y is positive, we take the positive square root:
y = 4√x
Now, substituting r = x and h = 4√x into the volume formula, we have:
V = ∫(2πx)(4√x)dx
Integrating this expression from x = 0 to x = 4, we can find the volume:
V = ∫(2πx)(4√x)dx
= 8π∫x^(3/2)dx
= 8π * [ (2/5)x^(5/2) ] evaluated from x = 0 to x = 4
= 8π * (2/5)(4^(5/2) - 0^(5/2))
Simplifying this expression further:
V = (16/5)π * (4^(5/2))
Calculating the value numerically:
V ≈ 815.72 cubic units
So, the volume generated when the area bounded by the curve y^2 = 16x, from x = 0 to x = 4, is rotated around the x-axis is approximately 815.72 cubic units.