Integration

Name: integration 5
Uploaded: 2022-01-04
Duration: 6 min 42 s
Channel: Numerade
Description: Integration

Question

Integration

Ekaveera Kumar · Accepted Answer

So this is our surface integral. We need to evaluate. So we need to use the divergence theorem. First of all, what is the region? The region is the region bounded by the parable Lloyd&#x27;s. That is going to X squared plus Y squared. And the plane, is it difficult to put And obviously the boundary of the surface will be circling. So you can see this diagram. So this is the diagram, this is the parable or is it is equal to X squared plus Y squared. This is my plane. That is equal to four. So if you can see this is a circle, X squared plus Y is critical before. So it&#x27;s better to use cylindrical coordinates to evaluate design divergence theorem. So what does divergence theorem says is the surface integral of a vector field? Dot ds D sigma is volume integral of the divergence of the vector field and the volume element D. V. So this is the whole human race of the triple integral. So let&#x27;s use cylindrical coordinates according to cylindrical coordinates We have accessible to our cost theater. Why are scientists and that is equal to that itself? And the jacobean of the cylindrical coordinates is well known to. First, I&#x27;m not doing that division to jacobins are. So that means first we need to set up the limits of the integration of the triple integral. And also we need to calculate the divergence of F. So first let&#x27;s calculate the divergence of F. So this is my vector field. So the divergence of F. The vector field and dr office going to the X component I call it as a fun divided by the X. So this is my F of X Y square. This is my F two. This is my everything. Let&#x27;s do F two by the way. Why Let&#x27;s do F three by doesn&#x27;t So what is the F one x 2? X is by square? What is do F two by Dubai? It is extra square. And what is the It is lucy Now when we substitute the cylindrical coordinates, the divergence becomes X squared plus y squared. In R squared enthusiasts as it is music. All right. So now so let us substituting now and if you see the limits of integration, Zet is depending on our because the parable Lloyd, Z is equal to X squared plus y squared. So the limits of our that belongs to 02. Our square I&#x27;ll use malted. Okay, I&#x27;ll use small it&#x27;s a small thing so small. That belongs to Ask her to 0 to ask her ask her to fall Because art is going from 0 to I will be going from 0-2 region. The boundaries a circle. The bounty is a circle of really spoke. So the equation of this circle is expert. Was very square is for the maximum radius is too the minimum radius is zero but that is actually depending on r squared. So that will be taking values from our square to four. So that is aspect for and r is zero to go. Of course 3 to 0 to two points because we have a complete revolution. So now we are ready to calculate them Surface integral. So the surface integral as after our basic model as he could do 3-0-2 pi Are 0-2 That are square to 4th. The divergence. How much we got? The divergence is R squared is 20. Let&#x27;s go ahead. Let&#x27;s do that to the Jacoby in art. D R. D data. So this is the order of integration suppose let us integrate with respect to them. Integration of that is equal to r squared forward, husband and wife is argue to reduce that integration of our cube with respect to Z. Is our cubes and integration of I can do this in order to do that. I&#x27;ll introduce you. What is the decoration of Brazilian? Does that square So it is ours that square. And the limits are low limit is sorry low limit is our square upper limit is four. So let&#x27;s substitute both the limits So it becomes our Cube into four. Art into 16 minus bracket are Cuban to our square. Our into our scoreboard square that is our powerful. So basically we&#x27;ll get for our cue for our cube plus 16 R minus r par fives are perfect. You are for five. So that means now let us spending your winter vision. It is theta is equal 0-5. R is equal to 0- two. For our cube Plus 16 R -2 are .5 D. R. Did he talk this integration? You can split it for R. And Theta because for the independent in degrees So theta is equal 02-5 digital into I is equal 0-2. For our cube Plus 16 R -2 are for five D. R. This is two because integration of the data is data And Parliament is zero. This integration let us do What is integration of hierarchy? Powerful integration of six. R 68 R squared integration into our power five. It is our power six divided by three substitute. It&#x27;s Caltech 2.2. Lower limit is anyway zero upper limit letters substitute to bar for 68 into 218-4 32 By 64 divided by three. So that will be 2.2 48 -64 divided by three. It is two pi by three and 2 18. So there&#x27;s 1 60 by by. So the answer for this surface intravenous 1 16 x 13.

Chris Trentman · Answer

were given a vector field f and the surface s and were asked to use the divergence theorem to calculate the surface Integral. Oprah s of f f is extra fourth, I minus execute C squared J plus four x y squared Z K and s is the surface of the solid talent bounded by the cylinder X squared plus y squared It was one and the planes Z equals X plus two and Z equals zero first, the divergence of F is or execute plus zero plus four x y squared. And by the divergence, the&#x27;re, um the surface integral over s of F is equal to the triple integral over the region e founded by S of the diversions of F, which is four x cubed plus four x y squared. And recognizing that our region E is symmetric about busy access, it makes sense to use cylindrical coordinates. Here. The projection on the X Y plane is a disk of radius one. And so we get the iterated integral, which is the integral from 0 to 2. Pi integral from R equals 01 integral from Z equals zero and Z equals X plus two, which in cylindrical coordinates is R CoSine data plus two of our function in cylindrical coordinates. This becomes for our cube cosign cube data plus four times are cubed cosine data times sine squared data. And in fact, we can combine this to get four are huge times co sign of data using triggered entities. And the differential is our Deasy DRD data taking the anti derivative with respect to Z and evaluating. We get integral from 0 to 2 pi integral from 01 of this is four are to the fourth co sign data times R cosine theta plus two. He already data and multiplying out and taking anti derivative with respect to our and evaluating you get the integral from 0 to 2 pi of let&#x27;s see Siri&#x27;s or yes, 4/6 or two thirds cosine squared of theta. Plus, this is 8/5 cosign data data and taking the anti derivative with respect to data and evaluating, we get well, we have two thirds times and then this will be one half times data which is two pi Yeah, which simplifies to two thirds pi

Chris Trentman · Answer

were given a vector field f and the surface s and were asked to use the divergence theorem to calculate the surface integral over s of f f is cosign Z plus x y squared I plus X e to the negative CJ plus sign y plus X squared zk and s is the surface of the solid founded by the Prabha Lloyd Z equals X squared plus y squared in the plane Z equals four. First of all, we have the divergence of f is why squared plus zero plus X squared and so Hi, detective Steel The surface integral over s of F is equal to the triple integral of the region bounded by s which I&#x27;ll just call e of the divergence of F which is X squared plus y squared disease and recognizing that E is symmetric about the Z axis, we make the switch to use cylindrical coordinates So the projection onto the X by plane is a disk of radius too. And so this becomes the iterated integral from 880 to 2 pi integral from r equals zero r equals two and integral from Z equals zero the X Y plane. We&#x27;re sorry. That&#x27;s equal zero from Z equals X squared plus y squared, which becomes R squared to Z equals four of our function X squared plus y squared which becomes R squared times the differential are dizzy DRD data taking the anti derivative with respect to Z and substituting Get the integral from 0 to 2 pi integral from 0 to 2 off our to fourth. Sorry, this should be just are this is our cue Times four minus r squared the rt data and then using Slovenia&#x27;s the&#x27;re, um we can right. This is a product of into girls. This is the integral from 0 to 2 pi the theater times the integral from 0 to 2 of or are cute minus are to the fifth D r and taking anti derivatives In evaluating we get two pi times are to the fourth minus 1/6 art of the sixth from zero to and plugging in Beit 32 3rd pie

Chris Trentman · Answer

were given a surface integral and were asked to evaluate it. This is the integral over S of X times Z Where s is thes surface which is the boundary of the region enclosed by the cylinder y squared plus C squared equals nine and the planes X equals zero and x plus y equals five mhm. So this is kind of a complicated surface. In fact, you could break us up into three surfaces. So we have the subsurface s one which is the lateral surface of the cylinder we have surface s to which is the front formed from the plane X plus y equals five and we have s three. And this is the back which is in the plain X equals zero. So on s one the surface is given by the vector equation. Well, since this is a cylinder are of UV equals you I plus the radius of the cylinder which is three cosign v j plus three Sign the okay where V ranges from zero to pi and we know that X. We don&#x27;t know much about you yet, but we know that X lies between zero and five minus why, which is the front plane. Now, rearranging this implies that you lies between still zero but then substituting five minus. And then why is in terms of U and V three cosign be so you was less than equal to five minus three co signed B Then we have that victor. Are you cross RV? Well, this is the vector one I plus you&#x27;re o J plus year. Okay, crossed with zero I minus three signed the J plus three cosign. Be okay And taking this we get negative. Three cosign v j minus three sign be que and of no coefficient for I and therefore the norm of are you crest with RB is going to be the square root of nine coastline squared V plus nine sine squared V which is the square root of nine or three and therefore we have that the integral over this subsurface s one of X Z is equal to the integral bringing this as an iterated integral, integral from V equals 0 to 2 pi integral from u equals zero to u equals five minus three co sign V of our function xz in terms of U and V. So this is going to be you. Times three sign V times the norm of are you cross RV which is three. Do you TV Taking the anti derivative With respect to you, we get factory at a nine times the integral from 0 to 2 pi and then we have one half you sign the sorry one half you squared sine V from 0 to 5, minus three cosign v d B and evaluating This is nine halves times the integral from 0 to 2 pi of five minus three cosign v squared times a sign of the TV and making a new substitution in your head. Yeah, or on paper. We get nine halves times and then we have one third so one over the coefficient in front of cosign v times one over the raised exponents, which is now three. So one third time&#x27;s one third or 1/9 times the expression five minus three co sign Be Now it&#x27;s cubed from 0 to 2 pi and evaluating. This is five See one half of Well, actually those were identical. So this is simply going to be zero. Another way you could have figured this out is to notice that this isn&#x27;t odd function. So that&#x27;s for S one now on s to while recall that this is the front of our region which has prioritization r of y z equals and we have the X is given by the equation of the plane five minus Why hi Plus why j plus z k. And we have because we&#x27;re constrained by the cylinder that why squared plus c squared must be less than or equal to nine. The equation of the cylinder and likewise we have that are y crossed with rz. This is simply the vector negative one I plus one j crossed with one K. This is simply I plus j and therefore the norm of our y Cross rz is the norm of this which is this squared of one plus one or route to and therefore the integral over the surface as to of X Z is what will substitute X in terms of why so this is now five minus y times z and then this is the double integral over Instead of y z such that y squared plus z squared It&#x27;s less than or equal to nine and then we also have in our into grand the norm of our way across RZ which is Route two d A. And we see that this region is circular and so we might use polar coordinates here. So making the switch, we get route two times the integral from 0 to 2 pi integral from r equals zero two r equals three of and then substitution gives us five minus R cosine data times are sign data and then the differential becomes r D r D theta Simplifying This is route two times the integral from 0 to 2 pi integral from 0 to 3 of five R squared minus R cubed Co signed data Times the sign of data D R D data taking the anti derivative With respect to our this becomes route two times integral from 0 to 2 pi of now this is five thirds are cubed minus 1/4 are to the fourth cosign data from 0 to 3. I&#x27;m signed data di Fada evaluating. We get route two times integral from 0 to 2 pi and this is 45 minus 81 4th co signed data times the sign of data D data This is equal to route to times and then factoring out four for 81 times and taking anti derivatives one half of the expression 45th minus 81 4th Cosine data now squared, evaluated from 0 to 2 pi. And if we plug in zero in two pi or really subtracting same term from itself And so this is going to be zero Finally, on the last part of the surface, the back as three. Well, we have the X is equal to zero. Therefore, it follows that the integral over the surface as three of x z. This is just going to be the integral of zero, which is zero and therefore putting this all together. The integral over s of xz is equal to zero plus zero plus zero for S one s two and s three, which is zero. So our answer

Chris Trentman · Answer

were given a surface integral. We were asked to evaluate it. This is three integral. Over the surface s of X squared Z squared. Where s is the part of the cone Z squared equals X squared plus y squared the lies between the planes Z equals one and Z equals three. So again we have the essence. The portion of the cones equals X squared plus y squared for Z between one and three. Equivalently s is part of the surface. Just still part of the cone c equals the positive square root of X squared plus y squared over the region in the X y plane d Just the set of pairs x y such that well, we need dizzy toe lie between zero one in three, which implies that X squared plus y squared also lies between one and between three squared since x squared plus y squared equals C squared, which is nine. Therefore, it follows that the surface integral over s of X squared Z squared is three double integral over the domain of integration. D um, X squared times and then Z squared is X squared plus y squared from our equation of our surface times. And then this is the square root of the partial derivative of Z with respect to X, which is if we multiply this out X over and square root of X squared plus y squared all squared, plus the partial derivatives e With respect to why? Which is why over the square root of X squared plus y squared squared plus one D A and simplifying be it. This is the double integral over de of we still have an X squared times X squared plus y squared and simplifying the inside of the radical. This is the square root of X squared plus y squared over X squared plus y squared plus one d A and this simplifies to double integral over de of route two times X squared times x squared plus y squared since neither well since X squared plus y squared is only going to be zero if both x and wire zero, which is impossible in our region. D. Since there&#x27;s some expert plus, Weisberg must be at least one now He can right. This is an iterated integral, so fact read a route to times integral and we see this is a region, which is easy to express in polar coordinates. So from beta equals zero to pi integral from r equals zero to r equals the radius of the disk, which is thes square root of nine or three and then in polar coordinates, X becomes R cosine theta an expert plus y squared becomes R squared and the differential becomes R D R D theta. And he&#x27;s in Fellini&#x27;s theory and weaken right. This is a product of minerals two times integral from 0 to 2 pi cosine squared data deep data times the integral from 0 to 3 of part of the fifth d r. Taking into derivatives, we get route two times and using trig identities or memorization, this is one half data plus 1/4 sign of two theta from 0 to 2 pi times 1/6 art of the sixth from 0 to 3. Substitution gives us route to times pie times 1/6. Sorry. That should be from R equals 123 Yeah, my mistake. Yeah. So times 1/6 of three to the sixth, minus one. And evaluating this simplifies 2 364 route to over three times pi, which is our answer

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