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.

Question

I am really confused on this question. I graphed all of the points, but I have no idea what to do.

Answer 1

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?

Answer 2

Yes I do see the relation, but I don't understand what to do next.

Answer 3

Huh? There's nothing to do next. I gave you the first integral. The 2nd one was just for extra credit.

Answer 4

Really that is all? I guess I was over complicating it. Thank you!