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In evaluating double integral over a region D, a sum of iterated integrals was obrained as follows:(1/3)y f(c,y) dxdy L % fle,y) drdy _IL f6,y)d4Sketch the region D...

Question

In evaluating double integral over a region D, a sum of iterated integrals was obrained as follows:(1/3)y f(c,y) dxdy L % fle,y) drdy _IL f6,y)d4Sketch the region D and express the double integral as an iterated integral with reversed order of integration.g(T) f(c,y) dydz 91 (z)91 (2)92

In evaluating double integral over a region D, a sum of iterated integrals was obrained as follows: (1/3)y f(c,y) dxdy L % fle,y) drdy _ IL f6,y)d4 Sketch the region D and express the double integral as an iterated integral with reversed order of integration. g(T) f(c,y) dydz 91 (z) 91 (2) 92



Answers

In evaluating a double integral over a region $ D $, a sum of iterated integrals was obtained as follows:
$$ \displaystyle \iint\limits_D f(x, y)\ dA = \int_0^1 \int_0^{2y} f(x, y)\ dx dy + \int_1^3 \int_0^{3 - y} f(x, y)\ dx dy $$
Sketch the region $ D $ and express the double integral as an iterated integral with reversed order of integration.

The question. We have privilege the civilian figure over two parts, a part viewing other the type off west Degen. Be part viewing on as a type of second region. No given double Integral is troubling. Figure over the regional exists where dear there are is a region bonded by bike ALS to 16 by X on Baikal stew X and X equals to it. So now, moving towards the solution, you can see figure from the question itself for a part since 80 argument by never reinvigoration work. That region on its spread is a close to invigoration from away and configuration from 16 by x. We're exist. Be right. Yes, we we're resolving the vigorous, so it would be keeping the outstanding figure that isn't situation. Wear exist where? Why limit going from 16 by X Who x Do you? Yes, they get limit from hope away. Extra power three minus 16 x The yes, which will be worth 5 70 six now moving towards might be since the area given by noble integration were the region are exist where the basic wants to our one last article that is, integration from full and integration from 16 bye Bye. Who laid excess? Where? The B by less integration from court away in good immigration. From why? Who? Weight. Excess Where? The X Day. By first, we will be smaller than the other. The pact. The village configuration from four one upon three X. You went from 16 by way. Who do you like less integration from 4 to 81 upon three to cover three. Why? Uh, we de By certain days we will get I seven physics. And this will be the final answer before they give a question. Thank you.

We want to write to give an integral with the order of integration. Change where that integral is integral. Zero. What integral? Zero X F X. Y. Dy dx. To change the order integration, we need to identify the appropriate balance because we can rewrite this integral as dx DY So we need to sketch this out to understand what those bounds are. I have a sketch here with the region that we're integrating over shaded in. So we see that changing the balance from 0 to 1 and 02 X requires us to think of how to invert the Y X coordinates. So we can write 0-1 as 0-1. Clearly, in terms of X. Y. As we can write Y X zero X zero. Y. I summarize these balancing on the right. So 01 becomes 0 to 10 X becomes zero Y. As observable on the figure on the left. That's a solution to this problem. Is the integral from 0 to 1 of the integral from zero to Y. F. Of X. Y. Dx Dy.

We want to write the given integral with the order of integration interchanged. The integral is the integral from one half to one of X cubed X F X. Y. Do I. D. S. This question challenge your understanding of multiple integral or indirect regular reaches of X. Y. Space. To appropriately change the order integration. We need to have dx dy or rather do I D X become the X. Y. With the balance changes appropriately. So what that means is by sketching and can identify the correct bounds. Mhm. We feel that the area in orange represents a given region we're integrating over. Thus we want to change our bounds from X. two Y numbers and we want to change our functions in terms of X two functions in terms of why? Yeah, we see that. This gives one half to one becoming 1/8 to 1 where 1/8 is Our x cubed at 1/2 X cubed X becomes wider than 30 Y. Thus are integral, is integral. 1/8 to 1 integral. Wide at 1 30 Y F X Y dx dy.

Is this problem? We want to write to give an integral with the order of integration interchanged. We have the integral from negative 120 of the integral From negative route Y plus one to root Y plus one. F. X. Y. The X Dy to change the order integration. We need our differentials to be in the form dy dx instead of the X. Dy with abound. Also changed appropriately to match this change in differentials. To identify the appropriate bound change to achieve this, we need to use a sketch of the graph to understand this. In the graph on the left we have the area shaded in orange representing the area where integrating over. We see that in terms of why coordinates negative 1-1, we can change them to X coordinates as negative 1-0. We can also change our bounds in terms of the function negative Y +12 Y plus one. By rewriting the functions. In terms of why infinite X that are bound I -1-0 Becomes negative 1-1, and our negative wire +12 Y plus one becomes x squared minus 1 to 0. Thus we have solution integral negative 1 to 1 integral experiment is 1 to 0. F X. Y Dy Dx.


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