Question

Consider the region bounded by the graphs of y = sin⁻¹(x), y = 0, and x = 1/2.b. Find the centroid of the region.

Accepted Answer

Identify the region bounded by the curves: the inverse sine function $$y = \sin$$^{-1}$$(x)$$, the line $$y = 0$$, and the vertical line $$x = \frac{1}{2}. $$This region lies between $$x = 0$$ and $$x = \frac{1}{2}$$ because $$\sin$$^{-1}$$(0) = 0$$ and the lower boundary is $$y=0. $$Set up the formulas for the coordinates of the centroid $$(\bar{x}, \bar{y}) of a $$planar region bounded by curves. The centroid coordinates are given by:

$$\displaystyle \bar{x} = \frac{1}{A} \int_a^b x \cdot f(x) \, dx$$ and $$\displaystyle \bar{y} = \frac{1}{2A} \int_a^b [f(x)]^2 \, dx$$,

where $$f(x) = \sin$$^{-1}$$(x)$$, $$a=0$$, $$b=\frac{1}{2}$$, and $$A is $$the area of the region. Calculate the area $$A of $$the region using the integral:

$$\displaystyle A = \$$int_0$$^{\frac{1}{2}} \sin$$^{-1}$$(x) \, dx.

$$This integral represents the area under the curve $$y = \sin$$^{-1}$$(x)$$ from $$x=0 to$$ $$x=\frac{1}{2}$$ above the $$x-$$axis. Compute the integrals needed for the centroid coordinates:

- For $$\bar{x}$$, evaluate $$\$$int_0$$^{\frac{1}{2}} x \sin$$^{-1}$$(x) \, dx.

- $$For $$\bar{y}$$, evaluate $$\$$int_0$$^{\frac{1}{2}} (\sin$$^{-1}$$(x))^2$$ \, dx$$.

These integrals may require integration by parts or substitution techniques. Finally, substitute the values of the integrals and the area A$ into the centroid formulas:

$$\displaystyle \bar{x} = \frac{1}{A} \$$int_0$$^{\frac{1}{2}} x \$$sin^{-1}(x$$) \, dx$$ and $$\displaystyle \bar{y} = \frac{1}{2A} \$$int_0$$^{\frac{1}{2}} (\$$sin^{-1}(x))^2$ \, dx.

$$This will give the coordinates of the centroid of the region.

Consider the region bounded by the graphs of y = sin⁻¹(x), y = 0, and x = 1/2.
b. Find the centroid of the region.

Risposta video verificata per un problema simile:

Concetti chiave

Inverse Sine Function (arcsin)

Centroid of a Plane Region

Definite Integrals for Area and Moments