Find the area of the region bounded by the curves y = sin^-1(x/6), y = 0, and x = 6 obtained by integrating with respect to y. Please include the definite integral and anti-derivative.
To find the area of the region bounded by the curves y = sin⁻¹(x/6), y = 0, and x = 6 when integrating with respect to y, we need to express the equations in terms of x instead of y.
Let's start by considering the curve y = sin⁻¹(x/6) and finding its equation in terms of x.
To do this, we need to recall the definition of the inverse sine function. The inverse sine of a value is the angle whose sine is the given value. In other words, sin⁻¹(x) = y if and only if sin(y) = x. Therefore, for our equation y = sin⁻¹(x/6), we can rewrite it as sin(y) = x/6.
Next, we need to eliminate the inverse sine and solve for y. Taking the sine of both sides of the equation sin(y) = x/6, we get sin(sin⁻¹(x/6)) = sin(y), which simplifies to x/6 = sin(y).
Now, we have the equation x/6 = sin(y), but we want to express y = f(x). To do this, we isolate y by taking the inverse sine of both sides: sin⁻¹(x/6) = y.
The equation y = sin⁻¹(x/6) is now expressed as y = f(x).
The region bounded by the curves also has the boundaries y = 0 and x = 6.
To find the area using integration with respect to y, we need to reverse the roles of x and y. This means we need to express x in terms of y and integrate with respect to y.
To find x in terms of y, we can solve the equation y = sin⁻¹(x/6) for x. Transposing the 6 to the other side of the equation, we have x = 6*sin(y).
Let's denote the area we're interested in as A. To find A, we integrate the difference of the two curves with respect to y over the interval [0, sin⁻¹(6/6)]:
A = ∫[0, sin⁻¹(6/6)] [(6*sin(y)) - 0] dy
Simplifying, we have:
A = ∫[0, sin⁻¹(6/6)] 6sin(y) dy
To integrate 6sin(y) with respect to y, we use the antiderivative of sin(y), which is -cos(y):
A = -6∫[0, sin⁻¹(6/6)] cos(y) dy
Evaluating the definite integral, we have:
A = -6[cos(y)] evaluated from 0 to sin⁻¹(6/6)
Since cos(y) evaluated at sin⁻¹(6/6) is simply cos(sin⁻¹(6/6)), which simplifies to cos(π/2), we have:
A = -6[cos(π/2) - cos(0)]
Using the trigonometric values, we find:
A = -6[0 - 1]
Simplifying further:
A = -6(-1) = 6
Therefore, the area of the region bounded by the curves y = sin⁻¹(x/6), y = 0, and x = 6 obtained by integrating with respect to y is 6 square units.
To find the area of the region bounded by the curves y = sin^-1(x/6), y = 0, and x = 6 by integrating with respect to y, we need to set up the integral and find the anti-derivative.
First, let's graph the given curves to better understand the region:
The curve y = sin^-1(x/6) is a portion of the sine function, with the range limited to [0, π/2] and the x-values limited to [0, 6].
The curve y = 0 represents the x-axis, and the line x = 6 is a vertical line passing through x = 6.
The area we want to find is bounded between y = sin^-1(x/6) and the x-axis (y = 0) within the x-range of [0, 6].
To find the limits of integration, we need to determine the points of intersection between y = sin^-1(x/6) and y = 0.
Setting y = sin^-1(x/6) equal to 0, we have:
0 = sin^-1(x/6)
Taking the sine of both sides, we get:
sin(0) = x/6
Simplifying, we find:
x = 0
Therefore, the lower limit of integration for y will be 0.
To find the upper limit of integration, we need to find the point of intersection between y = sin^-1(x/6) and x = 6.
Substituting x = 6 into y = sin^-1(x/6), we have:
y = sin^-1(6/6)
y = sin^-1(1)
Since sin^-1(1) = π/2, the upper limit of integration for y will be π/2.
Now we can set up the integral:
∫[0, π/2] (x) dy
To find the anti-derivative, we integrate x with respect to y:
∫ x dy = xy
Now, we can evaluate the definite integral:
Area = [xy] [0, π/2]
Substituting the limits of integration, we have:
Area = [x(π/2) - x(0)] [0, π/2]
Simplifying, we find:
Area = [6(π/2) - 6(0)] [0, π/2]
Area = [3π] [0, π/2]
Therefore, the area of the region bounded by the curves y = sin^-1(x/6), y = 0, and x = 6 is 3π square units.
y = sin^-1(x/6)
x = 6 siny
The boundaries define one-half of an arch on the curve, so the area is
∫[0,π/2] 6siny dy = 6