find volumes of solids generated by revolving the regions bounded by the graphs of the equations about the given lines.
y=square root of x
y=0
x=3
when the line x = 9 please
The region bounded by
y=√x
y=0
x=3
has its "vertices" at (0,0), (3,0), and (3,√3)
To rotate it about the line x=9, it's best to think of flat discs with holes in them. So, since the volume of a thin disc of thickness dy is πr^2 dy, and we need to subtract the holes, we have
v = ∫[0,√3] π(R^2-r^2) dy
where R = 9-x and r = 6
That makes
v = ∫[0,√3] π((9-y^2)^2-6^2) dy
= π(y^5/5 - 6y^3 + 45y) [0,√3]
= 144/5 √3 π
You can do it with thin shells of thickness dx, as well. The volume of such a shell is 2πrh dx, so
v = ∫[0,3] 2πrh dx
where r = 9-x h = y
v = ∫[0,3] 2π(9-x)√x dx
= 4/5 π (15-x) x^(3/2) [0,3]
= 144/5 √3 π
To find the volumes of the solids generated by revolving the region bounded by the graphs of the equations about the given lines, we can use the method of cylindrical shells.
The region bounded by the graphs of the equations y = √x, y = 0, and x = 3 can be visualized as a triangle with a curved hypotenuse.
First, we need to determine the limits of integration for x.
To find the x-values where the two curves intersect, we set y = 0 and y = √x equal to each other:
0 = √x
Squaring both sides, we get:
0 = x
Therefore, the region of interest is bounded by x = 0 and x = 3.
To set up the integral for calculating the volume, we will integrate with respect to x.
Let's consider a sample shell with height ∆x and radius r.
The height of each shell, ∆x, will be defined by the difference between the two curves at x.
The radius, r, will be the distance from the axis of rotation (x = 3).
Thus, the integral for calculating the volume V of the solid is:
V = ∫[0, 3] [(2πrh)dx]
To find r, we subtract the axis of rotation (x = 3) from the x-value:
r = x - 3
Now, let's substitute √x and 0 for h and integrate:
V = ∫[0, 3] [(2π(x - 3)√x)dx]
V = 2π ∫[0, 3] [x√x - 3√x]dx
Now, we can integrate term by term:
V = 2π [∫[0, 3] [x^(3/2)dx] - 3∫[0, 3] [√x dx]]
The first integral can be calculated using the power rule for integration:
∫[0, 3] [x^(3/2)dx] = [2/5x^(5/2)] from 0 to 3
Simplifying, we get:
(2/5)(3^(5/2)) - (2/5)(0^(5/2))
The second integral can also be calculated using the power rule:
-3∫[0, 3] [√x dx] = -3[2/3x^(3/2)] from 0 to 3
Simplifying further, we get:
-2(3^(3/2) - 0)
Substituting these values back into the volume equation:
V = 2π[(2/5)(3^(5/2)) - 2(3^(3/2))]
Simplifying the expression inside the brackets and multiplying by 2π:
V = 2π[(6/5)(3^(5/2)) - 2(3^(3/2))]
Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations y = √x, y = 0, and x = 3 about the line x = 3 is 2π[(6/5)(3^(5/2)) - 2(3^(3/2))].
To find the volumes of the solids generated by revolving the region bounded by the graphs of the equations about the given lines, we will use the method of cylindrical shells.
Given:
Equations: y = √x, y = 0, x = 3
Line of revolution: x-axis (y = 0)
Step 1: Graph the equations.
We can start by graphing the region bounded by the equations y = √x, y = 0 (x-axis), and x = 3. This will help us visualize the shape of the solid.
Step 2: Determine the limits of integration.
Since we are revolving the given region about the x-axis, the limits of integration will be the x-values that define the region. In this case, the region is bounded by y = √x, y = 0, and x = 3.
To find the limits of integration, we need to solve the equations y = √x and y = 0 for x.
Setting y = √x and y = 0, we get:
√x = 0 → x = 0 (lower limit of integration)
√x = 0 → x = 3 (upper limit of integration)
Therefore, the limits of integration are from x = 0 to x = 3.
Step 3: Setup the integral for the volume.
The volume of a solid generated by revolving the region bounded by the given equations can be expressed as an integral:
V = ∫[a to b] 2π * R(x) * h(x) dx
where R(x) is the radius of the shell and h(x) is the height of the shell.
In this case, the radius R(x) is the distance between the x-axis and the curve y = √x, which is given by R(x) = √x. The height h(x) is the infinitesimal thickness of the shell, which is Δx.
So, the integral to find the volume becomes:
V = ∫[0 to 3] 2π * √x * Δx
Step 4: Evaluate the integral.
To find the exact volume, we can evaluate the integral:
V = ∫[0 to 3] 2π * √x * dx
Integrating this with respect to x, we get:
V = 2π * ∫[0 to 3] x^(1/2) dx
V = 2π * [2/3 * x^(3/2)] | from 0 to 3
V = 2π * (2/3 * 3^(3/2))
Simplifying further:
V = 2π * (2/3 * 3 * √3)
V = 4π * √3
Step 5: Find the final answer.
The volume of the solid generated by revolving the given region about the x-axis is 4π√3 cubic units.
So, that's how you find the volume of solids by revolving regions bounded by equations!