{"id":432,"date":"2025-02-13T19:44:52","date_gmt":"2025-02-13T19:44:52","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/determining-volumes-by-slicing-learn-it-2\/"},"modified":"2025-02-13T19:44:52","modified_gmt":"2025-02-13T19:44:52","slug":"determining-volumes-by-slicing-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/determining-volumes-by-slicing-learn-it-2\/","title":{"raw":"Determining Volumes by Slicing: Learn It 2","rendered":"Determining Volumes by Slicing: Learn It 2"},"content":{"raw":"\n

Solids of Revolution and the Slicing Method<\/h2>\n

If a region in a plane is revolved around a line in that plane, the resulting solid is called a solid of revolution<\/strong>.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"529\"]\"This Figure 5. (a) This is the region that is revolved around the x-axis. (b) As the region begins to revolve around the axis, it sweeps out a solid of revolution. (c) This is the solid that results when the revolution is complete.[\/caption]\n\n

Solids of revolution are common in mechanical applications, such as machine parts produced by a lathe. We spend the rest of this section looking at solids of this type.<\/p>\n

The next example uses the slicing method to calculate the volume of a solid of revolution.<\/p>\n

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Use the slicing method to find the volume of the solid of revolution bounded by the graphs of [latex]f(x)={x}^{2}-4x+5,x=1,\\text{ and }x=4,[\/latex] and rotated about the [latex]x\\text{-axis}\\text{.}[\/latex]<\/p>\n\n[reveal-answer q=\"fs-id1167793998058\"]Show Solution[\/reveal-answer]
\n[hidden-answer a=\"fs-id1167793998058\"]\n\n

Using the problem-solving strategy, we first sketch the graph of the quadratic function over the interval [latex]\\left[1,4\\right][\/latex] as shown in the following figure.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"304\"]\"This Figure 6. A region used to produce a solid of revolution.[\/caption]\n\n

Next, revolve the region around the [latex]x[\/latex]-axis, as shown in the following figure.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"740\"]\"This Figure 7. Two views, (a) and (b), of the solid of revolution produced by revolving the region in (Figure) about the [latex]x\\text{-axis}\\text{.}[\/latex][\/caption]\n\n

Since the solid was formed by revolving the region around the [latex]x\\text{-axis,}[\/latex] the cross-sections are circles (step 1). The area of the cross-section, then, is the area of a circle, and the radius of the circle is given by [latex]f(x).[\/latex]<\/p>\n

Use the formula for the area of the circle:<\/p>\n

[latex]A(x)=\\pi {r}^{2}=\\pi {\\left[f(x)\\right]}^{2}=\\pi {({x}^{2}-4x+5)}^{2}[\/latex]  (step 2)<\/div>\n

The volume, then, is (step 3):<\/p>\n

[latex]\\begin{array}{cc}\\hfill V& =\\underset{a}{\\overset{h}{\\displaystyle\\int }}A(x)dx\\hfill \\\\ & ={\\displaystyle\\int }_{1}^{4}\\pi {({x}^{2}-4x+5)}^{2}dx=\\pi {\\displaystyle\\int }_{1}^{4}({x}^{4}-8{x}^{3}+26{x}^{2}-40x+25)dx\\hfill \\\\ & ={\\pi (\\frac{{x}^{5}}{5}-2{x}^{4}+\\frac{26{x}^{3}}{3}-20{x}^{2}+25x)|}_{1}^{4}=\\frac{78}{5}\\pi .\\hfill \\end{array}[\/latex]<\/div>\n

The volume is [latex]\\frac{78\\pi}{5}.[\/latex]<\/p>\n

[\/hidden-answer]<\/p>\n<\/section>\n

\n

Use the method of slicing to find the volume of the solid of revolution formed by revolving the region between the graph of the function [latex]f(x)=\\frac{1}{x}[\/latex] and the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[1,2\\right][\/latex] around the [latex]x\\text{-axis}\\text{.}[\/latex] See the following figure.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"523\"]\"This Figure 8.[\/caption]\n\n

[reveal-answer q=\"fs-id1167793261323\"]Show Solution[\/reveal-answer]
\n[hidden-answer a=\"fs-id1167793261323\"]<\/p>\n

[latex]\\frac{\\pi }{2}[\/latex]<\/p>\n

Watch the following video to see the worked solution to this example.<\/p>\n