Question

Center of gravity: Find the center of gravity of the region bounded by the x-axis, the curve y = sec x, and the lines x = -pi/4 and x = pi/4.

Accepted Answer

Identify the region bounded by the x-axis, the curve $$y = \sec x$$, and the vertical lines $$x = -\frac{\pi}{4}$$ and $$x = \frac{\pi}{4}. $$This region lies above the x-axis since $$\sec x is $$positive in this interval. Recall that the center of gravity (centroid) $$(\bar{x}, \bar{y}) of a $$planar region can be found using the formulas:  
$$\displaystyle \bar{x} = \frac{1}{A} \int_a^b x f(x) \, dx$$  
and  
$$\displaystyle \bar{y} = \frac{1}{2A} \int_a^b [f(x)]^2 \, dx$$,  
where $$A is $$the area of the region bounded by $$y = f(x)$$ and the x-axis from $$x = a to$$ $$x = b. $$Calculate the area $$A of $$the region using the integral:  
$$\displaystyle A = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec x \, dx.  
$$This integral represents the total area under the curve $$y = \sec x$$ between the given limits. Compute the $$x-$$coordinate of the centroid using:  
$$\displaystyle \bar{x} = \frac{1}{A} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} x \sec x \, dx.  
$$Note that because the function $$\sec x is $$even and $$x is $$odd, this integral will have symmetry properties that can simplify the calculation. Compute the $$y-$$coordinate of the centroid using:  
$$\displaystyle \bar{y} = \frac{1}{2A} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sec x)^2$ \, dx.  
$$This integral gives the moment about the x-axis, which when divided by twice the area, yields the vertical coordinate of the center of gravity.

Center of gravity: Find the center of gravity of the region bounded by the x-axis, the curve y = sec x, and the lines x = -pi/4 and x = pi/4.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Center of Gravity (Centroid) of a Region

Definite Integrals for Area and Moments

Properties and Behavior of the Secant Function