Question

39–44. Shell method about other lines Let R be the region bounded by y = x²,x=1, and y=0. Use the shell method to find the volume of the solid generated when R is revolved about the following lines.

y = 2

Accepted Answer

First, identify the region R bounded by the curves: $$y = x$$^{2}$$, x = 1$, and y = 0$. This region lies between x=0$ and x=1$ (since y$$=$$x^{2}$$ and $$y=0$$ intersect at $$x=0$$), and between $$y=0$$ and $$y=1 ($$since at $$x=1$$, $$y=1). $$Since the solid is generated by revolving the region R about the line $$y=2$$, which is horizontal and above the region, we use the shell method with vertical shells. The shells will be vertical slices parallel to the $$y-$$axis, so the variable of integration is $$x. $$The radius of a typical shell is the vertical distance from the shell at $$x to $$the axis of rotation $$y=2. $$Since the shell extends from $$y=0 up to$$ $$y=x$$^{2}$$, the radius is 2 - y$, but because the shell is vertical at position x$, the radius is 2 - y$ evaluated at the shell's height. For the shell method, the radius is the distance from the axis to the shell, so here it is 2 - y$, but since the shell height is y$$ = $$x^{2}$$, the radius is $$2 - x$$^{2}$$. The height of the shell is the horizontal length of the shell at position x$, which is from x=0$ to x=1$, so the height is simply x$-dependent. However, since the region is bounded by y=0$ and y$$=$$x^{2}$$, the height of the shell is the vertical distance between these curves, which is $$x^{2} - 0$$ = $$x^{2}. $$But for the shell method revolving around a horizontal line, the height corresponds to the horizontal length of the shell, which is the difference in $$x$$ values. Here, the shell is at position $$x$$, so the height is the vertical length of the shell, which is $$x^{2} - 0$$ = $$x^{2}. $$Actually, for the shell method about a horizontal line, the shells are vertical, so the height is the horizontal length of the region at that $$y. $$Since the region is described in terms of $$x$$, it's easier to integrate with respect to $$x$$, and the height is $$x ($$distance along $$x). $$The volume of the solid is given by the integral using the shell method formula: $$V = 2\pi \int_{a}^{b} (	ext{radius})(	ext{height}) \, dx. $$Here, $$a=0$$ and $$b=1$$, radius $$= 2 - x$$^{2}$$, and height = x$. So, set up the integral as V = 2$$\pi \int_{0}^{1} (2 - $$x^{2}$$) \cdot x \, dx$$.

39–44. Shell method about other lines Let R be the region bounded by y = x²,x=1, and y=0. Use the shell method to find the volume of the solid generated when R is revolved about the following lines.

y = 2

Verified video answer for a similar problem:

Key Concepts

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Setting up the Shell Radius and Height

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39–44. Shell method about other lines Let R be the region bounded by y = x²,x=1, and y=0. Use the shell method to find the volume of the solid generated when R is revolved about the following lines.y = 2

Verified video answer for a similar problem:

Key Concepts

Shell Method for Volume

Setting up the Shell Radius and Height

Region and Boundaries Identification

39–44. Shell method about other lines Let R be the region bounded by y = x²,x=1, and y=0. Use the shell method to find the volume of the solid generated when R is revolved about the following lines.

y = 2