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Evaluate the iterated integral by converting to polar coordinates. a 0 0 2x2y dx dy − a2 − y2Evaluate the iterated integral by converting to polar coordinates. Integral(0,a)integral(-sqrt(a^2-y^2), 0) x^2y dxdy ns transfered the x^2y right into polar coordinates:x= rcos(theta) , y= rsin(theta)(rcos(theta)^2(rsin(theta))r drd(theta) but, i"m no sure exactly how to transform the integral boundaries into polar coordiantes. Typically the an initial one would be like 0-pi yet i"m not sure with the a gift there. The a is throwing me off.

Evaluate the iterated integral by converting to polar coordinates. Integral(0,a)integral(-sqrt(a^2-y^2), 0) x^2y dxdy i transfered the x^2y right into polar coordinates:x= rcos(theta) , y= rsin(theta)(rcos(theta)^2(rsin(theta))r drd(theta) but, i"m not sure exactly how to transform the integral limits into polar coordiantes. Generally the very first one would certainly be choose 0-pi but i"m not sure with the a being there. The a is throwing me off.

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Monster Legends Hack No Survey No Human Verification Or SurveyFirst, let"s get an idea of what the domain being combined over is because that the original function:D = {(x,y) | 0 x = -sqrt(a^2 - y^2) --> this is the left fifty percent of a circle with radius ay = 0 -> a --> top top the left half of the circle, this is the top portionFrom these, I deserve to conclude that us are taking care of 1/4 of a circle. To be precise, we are handling the 1/4 of the circle that is in the second quadrant. From this, I deserve to conclude the the polar domain becomes:D = {(r,theta) | 0 can you do the trouble from here?

Ok, so below is what i got: int(0,a)int(pi/2, pi) (r^2cos(theta)^2)(rsin(theta))int(0,a)r^4 dr = r^5/5 <0,a> = a^5/5int(pi/2, pi) (cos(theta)^2)(sin(theta)) = -1/3cos(theta)^3

= 1/31/3 * a^5/5= a^5/15 Is the correct?I wasn"t sure if r^4 is correct. I was separating the r"s indigenous the cos and also sin, and merged with the other r, i got r^4.And i likewise wasn"t certain if the integration of (cos(theta)^2)(sin(theta)) was correct.

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