Exercise double integrals over rectangles evaluate the...

Exercise: Double Integrals over Rectangles Evaluate the double integral over the rectangular region R: R = {(x,y): 0 < x < 2, 0 < y < 3} âˆ¬(6xy + x) dA; R - {(x,y): 0 â‰¤ x â‰¤ 2, 0 â‰¤ y â‰¤ 3} âˆ¬(y + 2) dy; R = {(x,y): 0 < x < 2, 0 â‰¤ y â‰¤ 3} Evaluate âˆ¬(6xy + x) dA where: f(x,y) = 12xy - 8x; R = {(x,y): -1 â‰¤ x â‰¤ 2, 1 â‰¤ y â‰¤ 2} f(y) = y + 2; R is a closed rectangular region with vertices (-1,1), (2,1), (2,2), and (-1,2) f(x,y) = 4âˆš(x - 8y); R = {(x,y): y â‰¤ x â‰¤ 2y, 0 â‰¤ y â‰¤ 2}

Transcript

00:01 We need to evaluate the double integral integration of region r x square plus 4y cube dy d x the rectangle region basically we need to integrate over is x belongs to 1 to 2 and y belongs to 0 to 3 so basically we can do like this x is equal to 1 to 2 integration of y is equal to 0 to 3 x square plus 4 y cube divide so first we'll perform this integration with respect to y then with respect to x because it's at a angular region we can do like this so that will be integration of x is equal to 1 to 2 so what will be the integration of x square with respect to y because x square is a constant as plus y is concerned so what is integration of constant d y it is constant into y all right so y equal to 0 to 3 x squared d y integration is x square y.

01:09 So it is x square y and what is integration of 4 y cube? it is 4 y to the power 4 divided by 4.

01:16 Basically it's y to power 4.

01:18 So x square y plus y to power 4 now we need to substitute upper and lower limits 0 and 3 and this will be a function of y sorry function of x so we'll integrate with respect to x so let's substitute first upper limit so 3 x squared plus 3 power 4 81 but lower limit is 0s both are zeros so no issues d x now integrate with respect to x 3x square integration is 3x cube by 3 which is x cube and 81 integration with respect to x is 81 x now substitute upper and lower limits so upper limit is 2 cube which is 8 plus 81 into 262 and lower limit 1 cube which is 1 is 81 so the answer will be 170 minus 82 170 minus 82 is 88 so 88 is the answer for this question the volume above the region the tactical region is actually given by the double integral or the region are the surface z so what is the surface z we are given it is x root x square plus y and this will be d a the area element.

02:34 Now since the region is rectangular region, y belongs to 0 to 1 and x belongs to 0 to 1.

02:42 So we can integrate something like this.

02:46 Y is equal to 0 to 1.

02:48 Integration of x is equal to 0 to 1, x root x squared plus y d x and then we'll integrate with respect to y.

02:57 So first i'm doing with respect to x because it's very easy to integrate this with respect to x by using the substitution.

03:03 Let x square plus y is equal to u.

03:06 So when x square plus y is equal to you, the limits will change to y to y to x plus one because when x is zero this is y, when x is one, this is y plus one.

03:15 So it will be integration of y equal to zero to one...

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