For the iterated integral ??¹ ???² f(x,y) dx dy (sketch the...

For the iterated integral ??¹ ???² f(x,y) dx dy (sketch the region of integration and) write an equivalent integral with the order of integration reversed. (Note that ?y denotes the square root of y, and similarly for ?x. There are 9 possible answers listed below, multiple choice.) ??¹ ???¹ f(x,y) dy dx ??¹ ???² f(x,y) dy dx none of the other answers ??¹ ???? f(x,y) dy dx ??¹ ???? f(x,y) dy dx ??¹ ???¹ f(x,y) dy dx ??¹ ???² f(x,y) dy dx ??¹ ??²¹ f(x,y) dy dx ??¹ ??²¹ f(x,y) dy dx

Transcript

00:01 For this iterated integral, that is integral from 0 to 1 and integral from 0 to y squared, f of xy, dx, d, y.

00:10 We do the sketch of this iterated integral and also rewrite this after changing the order of integration.

00:16 So first, let's observe this iterated integral.

00:20 The first integral is with respect to x because we have dx here.

00:24 And this means for the first integral, these are the limits of x.

00:27 So we can write this as x equals 0 and x equals y squared and using this information we do the sketch of this region or this curve because when x equal to 0 basically represents the y axis and when x equal to y squared this represents this curve this is the curve represented for x equal to y squared now if we consider the outer integral that is with respect to y because we have d -y here and this means that these are the limits of y.

00:58 So therefore we write this as y equal to 0 or y equals 1.

01:02 So basically y varies from 0 to 1.

01:04 So here i have given the sketch of this region and so we have this curve and we also have the line y equals 1.

01:12 So first we take the rectangular slice of this area so that x varies from 0 to y square so that it touches this curve and then we take this rectangular slice of area from the bottom of this region that is from x from y equal to zero to y equal to one and that's why we got this iterated integral and if you consider after changing the order of integration so let's work on the sketch after we do the change of order of integration and for that we have to consider the vertical slice of this region and so when we consider the vertical slice of the region here y varies from this curve to this line so so for this curve, y equals root x because we can solve for y from this equation, y equals square root of x.

02:04 This is the lower limit of the first integral.

02:06 And then the upper limit of y will be it changes till this line.

02:11 That is it touches this line.

02:12 Y equals 1.

02:13 So the limits of y is from root x to 1.

02:18 And then we move this rectangle, this vertical slice of this area from this portion.

02:24 That is from x equal to 0 till this x equal to 1 because the point of intersection of the curve and the line is 1 .1...

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