Question

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.y=ln x,y=ln $$x^2$$; and y=ln 8; about the y-axis

Accepted Answer

First, identify the region R bounded by the curves: $$y = \ln x$$, $$y = \ln x^2$, and y$$ = \ln 8$$. Rewrite y$$ = \ln $$x^2 as$$ $$y = 2 \ln x to $$better understand the boundaries. Next, express the boundaries in terms of $$x$$ for a given $$y. $$From $$y = \ln x$$, solve for $$x to $$get $$x = e^y. $$From $$y = 2 \ln x$$, rewrite as $$y = \ln x^2$ and solve for x$ to get x$$ = $$e^{y/2}. $$The horizontal boundary is $$y = \ln 8. $$Determine the interval for $$y$$ over which the region exists. Since $$y$$ ranges from the lower intersection point of the curves up to $$y = \ln 8$$, find the intersection points by setting $$\ln x = 2 \ln x$$, which simplifies to $$\ln x = 0$$, so $$x=1$$, and find corresponding $$y$$ values. Set up the volume integral using the shell method since the region is revolved about the y-axis. The radius of a shell at position $$x is$$ $$x$$, and the height is the vertical distance between the curves in terms of $$x. $$The height is $$\ln 8 - \max(\ln x, 2 \ln x)$$ depending on the region. Write the volume integral as $$V = 2 \pi \int_{x=a}^{x=b} (	ext{radius})(	ext{height}) \, dx$$, where $$a$$ and $$b$$ are the $$x-$$values corresponding to the intersection points and the boundary $$y=\ln 8. $$Evaluate the integral to find the volume.

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.
y=ln x,y=ln x^2; and y=ln 8; about the y-axis

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Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.y=ln x,y=ln x^2; and y=ln 8; about the y-axis