5

Let G be the solid enclosed by the plane z =Y , the xy-plane, and the parabolic cylinder y= 1-x2 h(x) g(x,y) f{Jvav-f " { y dzdydx , then g(a,b)+h(4b) =a. -14b...

Question

Let G be the solid enclosed by the plane z =Y , the xy-plane, and the parabolic cylinder y= 1-x2 h(x) g(x,y) f{Jvav-f " { y dzdydx , then g(a,b)+h(4b) =a. -14b. -13~7d. -9-4f.-6g. ~2h -5

Let G be the solid enclosed by the plane z =Y , the xy-plane, and the parabolic cylinder y= 1-x2 h(x) g(x,y) f{Jvav-f " { y dzdydx , then g(a,b)+h(4b) = a. -14 b. -13 ~7 d. -9 -4 f.-6 g. ~2 h -5



Answers

Let $G$ be the solid in the first octant bounded by the sphere $x^{2}+y^{2}+z^{2}=4$ and the coordinate planes. Evaluate $$ \iint_{G} \int x y z d V $$ (a) using rectangular coordinates (b) using cylindrical coordinates (c) using spherical coordinates.

Want to find the volume under this plane and above this plane between these two parabolic cylinders. So first, let's draw the X Y plain. Let's draw two cylinders. We've got just a regular hold parabola and we've got this guy which is upside down upside down problem. So let's find these two intersection points. It's not too hard to see that this is when X squared is equal to one minus X squared. So two X is equal to X squared is equal to one. X squared is equal to one half X is equal to plus or minus one over the square root of two. So this is one over square root of two and negative one over square root of two and we're going to integrate with respects to y first. Okay, so looking at our double integral are expounds range. Think such are y bounds. Well, the lower ah function is, uh, this prevalent X squared so x squared and the upper one is one minus x squared and we see that we want to look at the plane. Are the salt under this plane and above this place. Hey, so looking at that, we've got three minus y few idea axe. Okay. And now we're going to just do some integration. Three. Why? Minus y squared over two from X squared, one minus X squared. Native one of her squared of two one over. Square root of two. These are three times one minus X squared, minus X squared. Squared over to minus. Um, sorry. Minus three times. Ah, sorry. I made a mistake back here. Ah, this should be one minus X squared squared over two nine ISS three times x squared minus X squared squared over two d X. All right, so, um, let's go ahead and combine these and see what we get. You've got three times one minus X squared, minus X squared. Right. That's combining these two guys here. Then we have Myers one minus x squared, squared over two plus X to the fourth over to DX. And now this is just a simple calculon into girl, right? We can combine like terms, et cetera. It's not going to be very eye opening for me to go through all the algebra, but we integrate. We played in our two bounds, and we will end up with five square root to over three

Fine line was the song by subtracting two volumes. So first thing way have this parabolic soldier and I'm going to draught in just two dimensions. So why equals X squared X Y planes? So we don't know what this is bounded by. So we need to figure out what it's bounded by, and we're gonna use these two planes to figure that out. So we're going to figure that out by setting these two planes equal to each other, and we can very, very easily verify that Y equals one is where these two planes intersect. So we have a plane y equals one here. Okay, so we've got these two intersections points one by one, and negative one one. Okay, so now we want to set up our double into girl. And the way that we're going to do that is we're going to integrate with respect, sto why and then X. So our expounds are negative one to one and our y bounds Well, the lower y bound is given by this curve here. This is our cylindrical parabolic cylinder. So are lower bound is X squared, and our upper bound is one of her bound is this guy in green here. Okay, so this is the region that we're looking at, and it is bounded above by this plane here and bounded below by this plan here again. Not too difficult to see that. So we're going to subtract d. Why the ex? So doing this Integration. We can see that we've got why? Well, let's let's write this out. We've got two eye minus two X squared one. Do idea, Jax from negative one to one of y squared minus two. Why? From X squared two one d x. So we've got we've got why square So we plug in one, get one minus two, Certainly plug in one And then when we plug in x the X to the fourth minus to x squared D x, I'm just verifying there already. So now we've got a negative one negative x to the fourth plus x squared d x. So this is a very easy into girl that we can figure out to dirt over X cubed for native one toe one right and plugging these guys in again. It's not very eye opening to do all this algebra. Our answer ends up being sixteen over fifteen

In English a back way have no stretch for lead in the first, hoping that encloses the plain X equals zero is that it was 20 and symbols to fight that minus y zero and 30 equals minus two y last six and be part way have to find the volume of this holder, but they can handle fact now moving. What must we will be going? The figure for this holy like this says X is going from 0 to 5. Why is going from zero? Also, why is going from the state is going from zero? Who then, to find the volume of the smaller breaking and why into the yet my nest White, less ex divi yet olden days. People first with Dana Arnau pull in signals integration, life, the express configuration from yet thanks for being with us before.

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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