Find the volume of the solid obtained when the region under the curve
y = 9 arcsin(x), x ≥ 0,
is rotated about the y-axis. (Use the Table of Integrals.)
Why did the curve go to the party? Because it wanted to get "a-round"!
To find the volume of the solid obtained by rotating the region under the curve y = 9 arcsin(x), x ≥ 0, about the y-axis, we can use the method of cylindrical shells.
Let's break it down step-by-step! First, we need to determine the limits of integration. The curve y = 9 arcsin(x) intersects the y-axis at y = 0, and the end point is at y = 9π/2. So our limits of integration are from y = 0 to y = 9π/2.
Next, we need to find the radius of each cylindrical shell. The radius at each y-value is given by x = sin(y/9).
Now, we can calculate the volume using the formula:
V = 2π∫[a,b] x(y)h(y) dy,
where x(y) is the radius and h(y) is the height of each cylindrical shell.
But wait, there's more! To solve the integral, we need to use the Table of Integrals or evaluate it numerically.
I hope this explanation made you smile while thinking about integrating and rotating curves! Let me know if you have any other questions or if you need further assistance.
To find the volume of the solid obtained when the region under the curve y = 9 arcsin(x), x ≥ 0, is rotated about the y-axis, we need to use the method of cylindrical shells.
The volume of the solid can be calculated using the formula:
V = 2π ∫[a, b] x * f(x) dx,
where a and b are the limits of integration corresponding to the region bounded by the curve.
In this case, we need to find the limits of integration, a and b, by solving for x in terms of y in the equation y = 9 arcsin(x).
Rearranging the equation, we have:
x = sin(y/9).
Now, we can set up the integral:
V = 2π ∫[a, b] x * f(x) dx
V = 2π ∫[a, b] sin(y/9) * y dy.
To determine the limits of integration, we need to find the values of y that correspond to the region under the curve. The function arcsin(x) has a range of [-π/2, π/2] and domain of [-1, 1]. Since the curve is y = 9 arcsin(x), the values of y will range from -9π/2 to 9π/2.
Therefore, the limits of integration are a = -9π/2 and b = 9π/2.
Now, we can calculate the volume using the integral expression:
V = 2π ∫[-9π/2, 9π/2] sin(y/9) * y dy.
To evaluate this integral, we may need to use the Table of Integrals or numerical methods, depending on the complexity of the integral.
To find the volume of the solid obtained when the region under the curve y = 9 arcsin(x) (for x ≥ 0) is rotated about the y-axis, we can use the method of cylindrical shells.
The formula for calculating the volume using cylindrical shells is given by V = ∫(2πy)(h)dy, where y represents the height of each shell, and h represents the thickness of each shell.
Step 1: Determine the limits of integration
To determine the limits of integration, we need to find the values of y for which the region under the curve y = 9 arcsin(x) exists. Since the curve is given for x ≥ 0, we can find the range of y values by solving the equation for x:
x = sin(y/9)
Since sin(y/9) has a range of [-1, 1], the y-values lie between -9 and 9.
Therefore, the limits of integration for y are from -9 to 9.
Step 2: Determine the height of each shell (y) and the thickness of each shell (h)
In this case, the height of each shell is given by y = 9 arcsin(x), and the thickness of each shell is represented by dy.
Step 3: Calculate the volume using the formula V = ∫(2πy)(h)dy
V = ∫[from -9 to 9] (2π)(9 arcsin(x))(dy)
Step 4: Solve the integral
To solve the integral, we can look it up in the Table of Integrals. Consulting the table, we find that the integral of arcsin(x) is x*arcsin(x) + sqrt(1-x^2) + C.
V = 2π ∫[from -9 to 9] (9x)(dy)
Since dy = dx, we can rewrite the integral as:
V = 2π ∫[from 0 to 9] (9x)(dx)
Simplifying,
V = 2π ∫[from 0 to 9] 9x dx
V = 2π * (9/2) * x^2 [from 0 to 9]
V = π * (9/2) * [9^2 - 0^2]
V = π * (9/2) * 81
V = 3645π cubic units
Therefore, the volume of the solid obtained when the region under the curve y = 9 arcsin(x) (for x ≥ 0) is rotated about the y-axis is 3645π cubic units.
I assume you mean only one period of the curve. If so, then using discs of thickness dy,
v = ∫[0,9π] πr^2 dy
where r = x = sin(y/9)
v = ∫[0,9π] π(sin(y/9))^2 dy = 9π^2/2
Using shells of thickness dx, and the symmetry of the region, we have
v = 2∫[0,1] 2πrh dx
where r = x and h = 9π/2 - y = 9π/2 - 9arcsin(x)
v = 2∫[0,1] 2πx(9π/2 - 9arcsin(x)) dx = 9π^2/2
integrate arcsin(x) using integration by parts