Question

20. Solid of revolution The region between the curve y=1/(2√x) and the x-axis from x=1/4 to x=4 is revolved about the x-axis to generate a solid.b. Find the centroid of the region.

Accepted Answer

Identify the region bounded by the curve $$y = \frac{1}{2\sqrt{x}}$$ and the x-axis from $$x = \frac{1}{4} to$$ $$x = 4. $$This region lies above the x-axis and below the curve within these limits. Recall that the centroid $$(\bar{x}, \bar{y}) of a $$planar region bounded by a curve and the x-axis 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, $$f(x) is $$the function defining the upper boundary, and $$[a,b] is $$the interval. Calculate the area $$A of $$the region using the integral:  
$$\displaystyle A = \int_{1/4}^4 \frac{1}{2\sqrt{x}} \, dx.  
$$This integral will give the total area under the curve from $$x=\frac{1}{4} to$$ $$x=4. $$Compute the $$x-$$coordinate of the centroid using:  
$$\displaystyle \bar{x} = \frac{1}{A} \int_{1/4}^4 x \cdot \frac{1}{2\sqrt{x}} \, dx = \frac{1}{A} \int_{1/4}^4 \frac{x}{2\sqrt{x}} \, dx.  
$$Simplify the integrand before integrating. Compute the $$y-$$coordinate of the centroid using:  
$$\displaystyle \bar{y} = \frac{1}{2A} \int_{1/4}^4 \left( \frac{1}{2\sqrt{x}} \$$right)^2$$ \, dx = \frac{1}{2A} \int_{1/4}^4 \frac{1}{4x} \, dx.  
$$Evaluate this integral to find $$\bar{y}.$$

20. Solid of revolution The region between the curve y=1/(2√x) and the x-axis from x=1/4 to x=4 is revolved about the x-axis to generate a solid.
b. Find the centroid of the region.

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Key Concepts

Solid of Revolution

Centroid of a Plane Region

Definite Integration for Area and Moments