Find the area of the region bounded by the curves of y=sin^-1(x/4), y=0, and x=4 obtained by integrating with respect to y. Your work must include the definite integral and the anti-derivative.
I am really confused on this question. I graphed all of the points, but I have no idea what to do.
The curve is maybe a bit easier to visualize if you write it as x = 4sin(y). The area is
∫[0,π/2] (4-x) dy
= ∫[0,π/2] (4-4sin(y)) dy = 2π-4
∫4-4sin(y) dy = 4y + 4cos(y)
This is a bit easier than using vertical strips, where the area is
∫[0,4] arcsin(x/4) dx
since integrating arcsin(x/4) involves integration by parts.
u = arcsin(x/4)
du = 1/√(16-x^2) dx
dv = dx
v = x
∫arcsin(x/4) dx
= x arcsin(x/4) - ∫x/√(16-x^2) dx
= x arcsin(x/4) + √(16-x^2)
You can see the relation between the two antiderivatives, right?
Yes I do see the relation, but I don't understand what to do next.
Huh? There's nothing to do next. I gave you the first integral. The 2nd one was just for extra credit.
Really that is all? I guess I was over complicating it. Thank you!
To find the area of the region bounded by the curves, we need to integrate with respect to y. However, since the given curves are not functions of y, we need to rewrite them in terms of y.
Let's start by looking at the curve y = sin^⁻¹(x/4). This equation gives us the y-coordinates of the points on the curve for a given x. To find the corresponding x-coordinates, we can take the sine of both sides:
sin(y) = x/4
Now we can express x in terms of y:
x = 4sin(y)
Next, let's consider the curve y = 0. Since y is always 0 for this curve, it represents the x-axis.
Lastly, the curve x = 4 is a vertical line passing through x = 4.
To find the boundaries of the region, we need to determine the maximum and minimum values of y. From the equation x = 4sin(y), we can see that the maximum value of y occurs when sin(y) is at its maximum value of 1, which happens when y = π/2. Similarly, the minimum value of y occurs when sin(y) is at its minimum value of -1, which happens when y = -π/2.
Now we can set up the definite integral to find the area. Since we are integrating with respect to y, the integral will have y as the variable of integration. We integrate from the minimum value of y to the maximum value:
Area = ∫[from -π/2 to π/2] [x(y) - 0] dy
Substituting the expression for x in terms of y, we have:
Area = ∫[from -π/2 to π/2] [4sin(y) - 0] dy
To evaluate this integral, we need to find the anti-derivative of 4sin(y), which is -4cos(y).
Area = [-4cos(y)] [from -π/2 to π/2]
Evaluating the anti-derivative at the limits, we get:
Area = [-4cos(π/2)] - [-4cos(-π/2)]
Simplifying, we have:
Area = [-4(0)] - [-4(0)] = 0
Therefore, the area of the region bounded by the curves y = sin^⁻¹(x/4), y = 0, and x = 4 is 0.