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Evaluate the triple integral over f(r,u,2) =IVz, where G is the solid in the first octant that i5 bounded by the parabolic cylinder z = 2 1? and the plancs z = 0,V ...

Question

Evaluate the triple integral over f(r,u,2) =IVz, where G is the solid in the first octant that i5 bounded by the parabolic cylinder z = 2 1? and the plancs z = 0,V = 2, and V =

Evaluate the triple integral over f(r,u,2) =IVz, where G is the solid in the first octant that i5 bounded by the parabolic cylinder z = 2 1? and the plancs z = 0,V = 2, and V =



Answers

Evaluate the triple integral. $\iiint_{G} x y z d V,$ where $G$ is the solid in the first octant that $G$ is bounded by the parabolic cylinder $z=2-x^{2}$ and the planes $z=0, y=x,$ and $y=0$

In the question we have to find the triple integral ex Davey. Where E is pounded by the parable light X equals to four by square plus four. Zero square and plain X equals to four. Now moving towards the solution here, uh let the big call to why Common said Why square plus? Is that square less than equal to one then triple integral over the region E X DV will be equal to double integral over the region D. Uh integration from four Y squared plus four. Zero square 24 X. D X. D. So this will be cool to double integral over the region day access square by two to the limit going from four by square plus four. That is square to four day A. Now moving further, it will be called to double integral over the region D eight minus eight into Y squared plus Z square D. A. So now using the polar coordinates we can right Integration from 0 to 2 pi into integration from 0 to 1. Eight minus eight are square to the power to R. D. R. D. Data. No solving this, you will get integration from 0 to 2 pi into integration from 0 to 18 minus eight R. To the power four R. D. R and D. Tita. So now moving further here you will get Integration from 0-2. 1st integrating in terms of far four, R squared minus four by three are to the past six. Integration from 0 to 1 day data. Now it will be equal to eight by three. Integration from 0 to 25 D three. To solving this, you will get eight by three and 2 to 5 which will be equal to 16 5 by three. And so this will be your answer. Thank you.

All right. So we want to value just triple into girl founded by the cylinder. And these planes were So we're only looking at the first often. So we're going to something a little weird and we're going to call. Why are sign gate our side, our coastline? Data and see, our scientists were switching it a little bit switching it up a little bit, but it doesn't matter, because these were just burial names so we can see that we've got the picture drawn over here. Okay, so there's our picture and we can see we're bounded like this. So our free well, maybe like to call our free variable is X. So if we look at our integration, we're going to see our bounds for axe here. Like so? Okay, bounce for our bounds for theta inbounds For X ray, we solve for y in terms of X. No resolved. We're x in. Why? Then we write out her into girl. Remember, this used to be easy but remember we call Z is equal to our scientific data and then we start to integrate Did treat me with respect to why we end up here integrating with respects too far. You're not going to do it all. It's a pretty simple integration. We end up with an eighty one who can pull that out. You're nice use of over here to get this. And then the rest is free. Straightforward to get twenty seven. Oh, great. As our answer.

We have triple integral. Why do you volume? We have equal Roy. And why equal one minus X s where? Therefore we have triple integral. Why D is a d y e x our boundaries for our negative one on one on for Why are zero and one minus X s where and forces are zero. And y know, integrate was suspected to X's it So we have double integral seeing boundaries. What is it? And our boundaries are zero on Loy de y dx. Now substitutes are money because we have double integral for the same boundaries. Boy square D Y E X now integrated with suspected Tuoi. So we have integral from negative 121 while over walk or three or three and our boundaries are zero and one minus x squared. The X now supposed to use our boundaries. So we have 1/3 integral from negative 1 to 11 minus x square. Hold for three weeks. Now make some possibility, Simplifications. We have 1/3. Integral off one minus three X square plus three x o bar four minus X power six the X. So we have 1/3 multiplied by X minus X Power three plus three or five export five minus exports saving over seven and our boundaries. Nick, the phone on one now subsidies our boundaries. So we have 32 over 105.


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