Fill in the blanks to set up the cylindrical integral...

Fill in the blanks to set up the cylindrical integral correctly to be equivalent to the cartesian integral:\\ $\int_0^6 \int_0^{6-x} \int_0^{9-x^2-y^2} \frac{y}{x} dzdydx = \int_0^{[[1]]} \int_0^{[[2]]} \int_0^{[[3]]} rdzdrd\theta$

Transcript

00:01 So you have the following integral in cylindrical coordinates, which are at least from 0 to 2 pi.

00:17 And from 0 up to 1, 1 between r and 1 over square root of 2 minus r squared.

00:37 We're square width of 2 minus r squared 3d z r d r d ther so this first bounds correspond to the bound on z the bound and this is obviously for there so for this first interval all the interval of z dc of 3 dc is equal to just 3 times z between the two bounds so that this will be equal to so 3 times this one c for the upper bound which is 1 over square root of 2 minus r squared minus 3 times the other bound are so that we have this r -tr between the next this is the next integral to do 0 and 1 and 0 to pi d theta so now all this integral can be separated into two intervals three r over square bit of two r squared and the interval of 3r squared well for this interval you can do a use substitution u equals to r squared u equals to r squared u equals to r squared so that this would be this integral 3 r d r t r squared square is of 2 my minus r squared while this is u and then the u would be equal to um to r d r so so that this would be equal to the integral of 3 over 2 because you move the d u half is equal to r d u over 2 d u 1 minus u so the integral of 1 minus u is equal to so 2 times 1 minus u and the way to check is well if we differentiate this we differentiate the right -hand side we should be at the left so if you differentiate that you have a minus 2 and then a half because this is here today one half and then that to the minus 1 half and then minus times a minus 1 which if you check is the same so this is the our interval and well this would be the integral of that so that putting it three halves so three halves of that is able to minus 3 halves and then squared that 0 .0b equal to minus 3 square root of 1 minus u and the r u is equal to r squared so that this interval the interval from 0 to 1 3 r d r r squared of 1 2 minus r squared will be equal to so we go to minus 3 times the square of 2 minus r squared but weight between 0 and 1 which is over 1 you have minus it was still 1 half minus minus 3 times 2 minus 1, 1 to 1 1 1 1 1 1.

06:58 And then at 0 minus 3 and then times the square root of 2.

07:13 So that this will be equal to 3 times the square root of 2 minus 1...

Question

Fill in the blanks to set up the cylindrical integral correctly to be equivalent to the cartesian integral:\\ $\int_0^6 \int_0^{6-x} \int_0^{9-x^2-y^2} \frac{y}{x} dzdydx = \int_0^{[[1]]} \int_0^{[[2]]} \int_0^{[[3]]} rdzdrd\theta$

Please give Ace some feedback