Question

69. Comparing volumes Let R be the region bounded by y = sin x and the x-axis on the interval [0, π]. Which is greater, the volume when R is revolved about the x-axis, or the volume when R is revolved about the y-axis?

Accepted Answer

First, identify the region R bounded by the curve $$y = \sin x$$ and the x-axis on the interval $$[0, \pi]. $$This means the region lies between $$y = 0$$ and $$y = \sin x$$ for $$x in$$ $$[0, \pi]. To $$find the volume when R is revolved about the x-axis, use the disk method. The volume $$V_x is $$given by the integral:

$$V_x = \pi \$$int_0$$^{\pi} (\sin x)^2$ \, dx$$

because the radius of each disk is $$\sin x. To $$find the volume when R is revolved about the y-axis, use the shell method. The volume $$V_y is $$given by the integral:

$$V_y = 2\pi \$$int_0$$^{\pi} x \cdot \sin x \, dx$$

where $$x is $$the radius of the shell and $$\sin x is $$the height. Set up both integrals explicitly and prepare to evaluate them (though we won't compute the final values here). This involves recognizing the integrals:

$$\$$int_0$$^{\pi} (\sin x)^2$ \, dx$$ and $$\$$int_0$$^{\pi} x \sin x \, dx. $$After evaluating both integrals, compare the numerical values of $$V_x$$ and $$V_y to $$determine which volume is greater.

69. Comparing volumes Let R be the region bounded by y = sin x and the x-axis on the interval [0, π]. Which is greater, the volume when R is revolved about the x-axis, or the volume when R is revolved about the y-axis?

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Key Concepts

Volume of Solids of Revolution

Disk/Washer Method

Shell Method