{"id":428,"date":"2025-02-13T19:44:50","date_gmt":"2025-02-13T19:44:50","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/areas-between-curves-learn-it-3\/"},"modified":"2025-02-13T19:44:50","modified_gmt":"2025-02-13T19:44:50","slug":"areas-between-curves-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/areas-between-curves-learn-it-3\/","title":{"raw":"Areas Between Curves: Learn It 3","rendered":"Areas Between Curves: Learn It 3"},"content":{"raw":"\n

Regions Defined with Respect to [latex]y[\/latex]<\/h2>\n

In the previous example, we had to evaluate two separate integrals to calculate the area of the region. However, there is another approach that requires only one integral. What if we treat the curves as functions of [latex]y,[\/latex] instead of as functions of [latex]x?[\/latex]<\/p>\n

Review the last example.<\/p>\n

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Consider the region depicted in the following figure. Find the area of [latex]R.[\/latex]<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"250\"]\"This Figure 7.[\/caption]\n<\/section>\n

Note that the left graph, shown in red, is represented by the function [latex]y=f(x)={x}^{2}.[\/latex] We could just as easily solve this for [latex]x[\/latex] and represent the curve by the function [latex]x=v(y)=\\sqrt{y}.[\/latex] However, based on the graph, it is clear we are interested in the positive square root. <\/p>\n

Similarly, the right graph is represented by the function [latex]y=g(x)=2-x,[\/latex] but could just as easily be represented by the function [latex]x=u(y)=2-y.[\/latex]<\/p>\n

When the graphs are represented as functions of [latex]y,[\/latex] we see the region is bounded on the left by the graph of one function and on the right by the graph of the other function. Therefore, if we integrate with respect to [latex]y,[\/latex] we need to evaluate only one integral.<\/p>\n

Let\u2019s develop a formula for this type of integration.<\/p>\n

Let [latex]u(y)[\/latex] and [latex]v(y)[\/latex] be continuous functions over an interval [latex]\\left[c,d\\right][\/latex] such that [latex]u(y)\\ge v(y)[\/latex] for all [latex]y\\in \\left[c,d\\right].[\/latex] We want to find the area between the graphs of the functions, as shown in the following figure.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"267\"]\"This Figure 8. We can find the area between the graphs of two functions, [latex]u(y)[\/latex] and [latex]v(y).[\/latex][\/caption]\n\n

This time, we are going to partition the interval on the [latex]y\\text{-axis}[\/latex] and use horizontal rectangles to approximate the area between the functions. So, for [latex]i=0,1,2\\text{,\u2026},n,[\/latex] let [latex]Q=\\left\\{{y}_{i}\\right\\}[\/latex] be a regular partition of [latex]\\left[c,d\\right].[\/latex] Then, for [latex]i=1,2\\text{,\u2026},n,[\/latex] choose a point [latex]{y}_{i}^{*}\\in \\left[{y}_{i-1},{y}_{i}\\right],[\/latex] then over each interval [latex]\\left[{y}_{i-1},{y}_{i}\\right][\/latex] construct a rectangle that extends horizontally from [latex]v({y}_{i}^{*})[\/latex] to [latex]u({y}_{i}^{*}).[\/latex] <\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"489\"]\"This Figure 9. (a) Approximating the area between the graphs of two functions, [latex]u(y)[\/latex] and [latex]v(y),[\/latex] with rectangles. (b) The area of a typical rectangle.[\/caption]\n\n

The height of each individual rectangle is [latex]\\text{\u0394}y[\/latex] and the width of each rectangle is [latex]u({y}_{i}^{*})-v({y}_{i}^{*}).[\/latex] Adding the areas of all the rectangles, we see that the area between the curves is approximated by:<\/p>\n

[latex]A\\approx \\underset{i=1}{\\overset{n}{\\text{\u2211}}}\\left[u({y}_{i}^{*})-v({y}_{i}^{*})\\right]\\text{\u0394}y.[\/latex]<\/div>\n

This is a Riemann sum, so we take the limit as [latex]n\\to \\infty ,[\/latex] and we get:<\/p>\n

[latex]A=\\underset{n\\to \\infty }{\\text{lim}}\\underset{i=1}{\\overset{n}{\\text{\u2211}}}\\left[u({y}_{i}^{*})-v({y}_{i}^{*})\\right]\\text{\u0394}y={\\displaystyle\\int }_{c}^{d}\\left[u(y)-v(y)\\right]dy.[\/latex]<\/div>\n

These findings are summarized in the following theorem.<\/p>\n

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finding the area between two curves, integrating along the [latex]y[\/latex]-axis<\/h3>\n

Let [latex]u(y)[\/latex] and [latex]v(y)[\/latex] be continuous functions such that [latex]u(y)\\ge v(y)[\/latex] for all [latex]y\\in \\left[c,d\\right].[\/latex]<\/p>\n

 <\/p>\n

Let [latex]R[\/latex] denote the region bounded on the right by the graph of [latex]u(y),[\/latex] on the left by the graph of [latex]v(y),[\/latex] and above and below by the lines [latex]y=d[\/latex] and [latex]y=c,[\/latex] respectively.<\/p>\n

 <\/p>\n

Then, the area of [latex]R[\/latex] is given by:<\/p>\n

[latex]A={\\displaystyle\\int }_{c}^{d}\\left[u(y)-v(y)\\right]dy.[\/latex]<\/div>\n<\/section>\n
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Back to our previous example, let\u2019s integrate with respect to [latex]y[\/latex]. Let [latex]R[\/latex] be the region depicted in the following figure. Find the area of [latex]R[\/latex] by integrating with respect to [latex]y.[\/latex]<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"250\"]\"This Figure 11.[\/caption]\n\n

[reveal-answer q=\"fs-id1167793589589\"]Show Solution[\/reveal-answer]
\n[hidden-answer a=\"fs-id1167793589589\"]<\/p>\n

[latex]\\frac{5}{3}[\/latex] units2<\/sup><\/p>\n

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