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July 23rd, 2017, 12:59 AM   #1
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Double integrals

Ok, so.. I have this double integral and I am asked to draw the area of integration (I hope this is how you call it in English) and change the order of integration. Now, I'm not sure do I take x=0, x=1/4, x=1 and y=1/2, y=1, y=sqrt(x) and draw them? Or? Attached Images 20289606_1313527688745340_312554787_n.jpg (37.5 KB, 5 views) July 23rd, 2017, 04:40 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 There are two double integrals here so two areas of integration. (Yes, that is how you say it in English- your English is excellent.) (If I wanted to be really, really, really "hard nosed" (strict) I would say that "area" is a number and you should use "region" of integration but even native speakers of English aren't that strict!) The first double integral is $\displaystyle \int_0^{\frac{1}{4}}dx \int_{\frac{1}{2}}^1 e^{\frac{x}{y^2}}dy$. (That is a perfectly valid way of writing a double integral though I would be more used to $\displaystyle \int_0^{\frac{1}{4}} \int_{\frac{1}{2}}^1 e^{\frac{x}{y^2}}dy dx$.) Determining that area. and changing the order of integration is particularly easy, almost trivial, for this particular integral because the limits of integration are all constants! x ranges from 0 to 1/4 for all y and y ranges from 1/2 to 1 for all x. The area of integration is the rectangle with the y-axis, x= 0, as left side, x= 1/4 as right side, y= 1/2 as bottom, and y= 1 as top. Changing the order of integration is just $\displaystyle \int_{\frac{1}{2}}^1 dy\int_0^{\frac{1}{4}} e^{\frac{x}{y^2}} dx$ (or $\displaystyle \int_{\frac{1}{2}}\int_0^{\frac{1}{4}} e^{\frac{x}{y^2}} dx dy$). The second integral is $\displaystyle \int_{\frac{1}{4}}^1 dx \int_{\sqrt{x}}^1 e^{\frac{x}{y^2}} dy$. That's just a little harder. x goes from $\displaystyle \frac{1}{4}$ to 1 for all y so draw the two vertical lines at x= 1/4 and x= 1. y goes from the curve $\displaystyle y= \sqrt{x}$ to 1. Draw the graph of $\displaystyle y= \sqrt{x}$ and the line y= 1. The area of integration is the area bounded by those four graphs. The curve $\displaystyle y= \sqrt{x}$ crosses the line x= 1/4 at (1/4, 1/2) and crosses the line x= 1 at (1, 1). The line y= 1 crosses the line x= 1/4 at (1/4, 1) and crosses the line (1, 1) at (1, 1). The change the order of integration, note that the lowest value of y in that region is 1/2 and the largest is 1. And, for every y, x goes from 1/4 on the left to $\displaystyle x= y^2$ on the right. The integral can be written $\displaystyle \int_{\frac{1}{2}}^1 dy\int_{\frac{1}{4}}^{y^2} e^{\frac{x}{y^2}} dx$. Thanks from sarajoveska Last edited by Country Boy; July 23rd, 2017 at 04:43 AM. July 23rd, 2017, 06:35 AM   #3
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Quote:
 Originally Posted by Country Boy There are two double integrals here so two areas of integration. (Yes, that is how you say it in English- your English is excellent.) (If I wanted to be really, really, really "hard nosed" (strict) I would say that "area" is a number and you should use "region" of integration but even native speakers of English aren't that strict!) The first double integral is $\displaystyle \int_0^{\frac{1}{4}}dx \int_{\frac{1}{2}}^1 e^{\frac{x}{y^2}}dy$. (That is a perfectly valid way of writing a double integral though I would be more used to $\displaystyle \int_0^{\frac{1}{4}} \int_{\frac{1}{2}}^1 e^{\frac{x}{y^2}}dy dx$.) Determining that area. and changing the order of integration is particularly easy, almost trivial, for this particular integral because the limits of integration are all constants! x ranges from 0 to 1/4 for all y and y ranges from 1/2 to 1 for all x. The area of integration is the rectangle with the y-axis, x= 0, as left side, x= 1/4 as right side, y= 1/2 as bottom, and y= 1 as top. Changing the order of integration is just $\displaystyle \int_{\frac{1}{2}}^1 dy\int_0^{\frac{1}{4}} e^{\frac{x}{y^2}} dx$ (or $\displaystyle \int_{\frac{1}{2}}\int_0^{\frac{1}{4}} e^{\frac{x}{y^2}} dx dy$). The second integral is $\displaystyle \int_{\frac{1}{4}}^1 dx \int_{\sqrt{x}}^1 e^{\frac{x}{y^2}} dy$. That's just a little harder. x goes from $\displaystyle \frac{1}{4}$ to 1 for all y so draw the two vertical lines at x= 1/4 and x= 1. y goes from the curve $\displaystyle y= \sqrt{x}$ to 1. Draw the graph of $\displaystyle y= \sqrt{x}$ and the line y= 1. The area of integration is the area bounded by those four graphs. The curve $\displaystyle y= \sqrt{x}$ crosses the line x= 1/4 at (1/4, 1/2) and crosses the line x= 1 at (1, 1). The line y= 1 crosses the line x= 1/4 at (1/4, 1) and crosses the line (1, 1) at (1, 1). The change the order of integration, note that the lowest value of y in that region is 1/2 and the largest is 1. And, for every y, x goes from 1/4 on the left to $\displaystyle x= y^2$ on the right. The integral can be written $\displaystyle \int_{\frac{1}{2}}^1 dy\int_{\frac{1}{4}}^{y^2} e^{\frac{x}{y^2}} dx$.
Thank you for this! I checked the solution in my book and it says that I need to get one region and I also have trouble solving that first integral because I get integral of y^2(e^(1/4y^2)-1). And I have no idea what to do with this. Tags double, integrals Search tags for this page

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