5

47. The region inside the curve 'cos 0 and inside the circle 1/V2 in the first quadrant...

Question

47. The region inside the curve 'cos 0 and inside the circle 1/V2 in the first quadrant

47. The region inside the curve 'cos 0 and inside the circle 1/V2 in the first quadrant



Answers

$23-28$ Find the area of the region that lies inside the first curve
and outside the second curve.

$$
r=3 \cos \theta, \quad r=1+\cos \theta
$$

So for this problem were asked to find the area that is inside both of the equations but is also in the first quadrant. And the two equations are R equals square root of co sign data and R equals one over the square root of two. So the first thing I'm gonna do is I'm going to graph r equals one over the square to two, which is about 20.7. And then the next thing I'm gonna do is I'm gonna find out the values of the square root of co sign of data. So coastline of zeros one square root of one is one. Then these co sign of data over to a zero score 200 co sign of pious, negative one. And unfortunately, there's not a square root of negative one, so that does not exist. The coastline of three pi over two is going to be zero square it a zero a zero, and then the coastline of two pi is going to be one square root of one is one. So I'm gonna go over here and graph this. So if you recall, we have zero hi over to, uh, hi. Three pi over two and then to pipe. So whenever so it data being zero, we have one and then we have this going to the zero at Pi over two. Hi doesn't exist. Therefore, three pi over to a zero and then it comes back around to want to play. So what we want to do is we want to find out the area inside both circles but Onley in quadrant one. So we're looking for this area here. So what we need is the intersection point that is right here. So in order to do that, we're gonna have to find out where these two equations equal each other. So are one equaling are too so square root of co sign of data equals one over the square root of two. If I square both sides, I get coastline data Square root of one has won. I'm sorry. One squared and then squirt of to squared is gonna be too so co sign equals 1/2 at high over three and at five pi over three. Now, looking back at the graph, we're like, I need one of these because we're only looking at one quadrant. So this intersection point is going to be pi over three. So we're gonna have to break this one up into two pieces. So looking at this, we have worth EDA is zero to so going from 0 to 2 pi over three for what's in green, which is R r equals one over squared too. So if we notice we're going from here to here, and so that's going to cover this portion of the red And in order to get the blue, we've got pi over three soup I over to because if you recall going from that zero over to you, hi, over to God is to our 1st 0 and that's going to take care of this little sliver of here. So many used the formula A equals 1/2 from B to a or a B of our squared Destatis. Since I have two pieces, I'm gonna have two parts to this. So the 1st 1 do 1/2 is I'm going to do the green section, which I said was from zero to pi over three and the green one is going to be our first or one of our equations, which is one over square root of two. So pie over 3 to 0 of one over square root of two squared dif ate up and then from hi over to two pi over three of our square root of co sign data squared defeat. So we go back up here we said the blue section which we were gonna add to it to make it all part of what was underneath that from pie itude of high three, I'm sure privately to Piper to you for square root of co sign data. So now minute, go ahead and take care of these squared. So one squared is one over squared it two squared is too, so I can combine these to make 1/4 and then anything under square root scored is just going to be itself. And I believe that half outside because we could bring that back in later. And now I can integrate. So when I have 1/4 data plus 1/2 of science data, the 1/4 is gonna be from PI over three 20 And the 1/2 sign is gonna be from pi over 32 pi over two. So for the first part, I'm gonna have high over three times 1/4. So it's gonna be pi over 12 minus zero because 1/4 time 00 and then high sign of pi over two is going to be one. So one times 1/2 and then minus sign over. Sign of pi over three is going to be the square root of 3/2 times. 1/2 that's gonna make this 1/4. Then I'm gonna put these all together someone I have pi over 12. Plus, I want to change this to 2/4 so that way I can combine these. So I've got pi over 12 plus two minus the square root of three all over for

So if the following um set of problems, we're going to be looking to find the area um within this polar curve, the region bounded by the polar curve and what this is going to be very common formula that we're gonna be using in this section. That is equal to the integral from A to B. Of 1/2 r squared. I'm going to move the one half outside and then we have one half art squared the fatah. So then what that's going to end up showing us is that we know that our is going to be equal to the function with respect to data. So we get our, in terms of data in this case stated squared. So r squared would actually be fae to to the fourth so we can plug this in and we get um it should have certain power and depending on the equation that were given, will plug that in and then we'll also change the bounds of integration to give us our final answer. But these are the two equations that will allow us to solve any polar equation occurs to find the area bounded by them.

What it will do. Synnex where x greater than an equal two. Xu X less than and equal to buy. So in discussion, we have to use a graph to give a rough estimate of the area of the region that lie beneath the Given Co. Then we have to find the exit area so you can see that here growth for this the growth, the function and mark the area of the region Beneath eight. We can determine the approximate value of the area given to write congruent triangle with legs 5.4 and 0.7 and two congruent preppers. Wait with base 0.7 and one and hide pipe on four. You can see that in this graph, so approximate where you is two point so it two times by a born fall going 2.7 upon two plus two times 0.7 plus one time 5.4 upon to approximate value is 0.550 plus 1.3 five Sorry, 35 So here approximate value is 1.885 Now here, compute the exit area. Intrigued Synnex li x zero to pi. So here we're not the Androgel of cynics equal to negative core sex. So we put here negative Cossacks. Zero to pi. Now we put the limit. Now you do cools. Why? Negative called seal. So now we know that the value of course, by his narrative now negative one. So we put here negative one negative calls. Your level is one. So we put here one and we get here Negative time. Negative one. Sorry to cool. No, you can see that here we get to Because negative times negative, equal to positive Soviet Soviet here. To which is positive. So it is all violence.

Problem. Number 30 r is equal to one plus co sign. Fator area is equal to integration from Miracle Point over to over half one plus co sign Fator Square The State Act, which is equal to integration from your apple pie over to off half one plus co sign it Squared stated Lost Who co signed data data which is equal to integration off ah one plus one plus co. Sign to theatre over two plus two. Core sign Peter Defeated, which is equal to have zero plus, uh, fainted. So the stater is it Data lost half a theater lost sign to with data over four plus to sign at a theater from zero toe by over two. Using substitution eso The final answer will be and a half by over two minus zero plus half by over two minus zero bus zero minus 0/4 plus toe one minus zero, which is equal to ah, half by over two plus boy over four plus two, which is equal to, uh, pretty three. A boy over a plus one


Similar Solved Questions

5 answers
Question 111Whatlis 4 (6* Hin x) ? dxIn 6 6" Inx+ 6"In 6"6x 1In 6 6* + 1x6'-1_ 6* In x +
Question 111 Whatlis 4 (6* Hin x) ? dx In 6 6" Inx+ 6" In 6"6x 1 In 6 6* + 1 x6'-1_ 6* In x +...
5 answers
Question 3 of 9 (1 point) Attempt 1 of UnlimitedSolve the absolute value equation M+s--5The solution set isCheck AnswerSeareh
Question 3 of 9 (1 point) Attempt 1 of Unlimited Solve the absolute value equation M+s--5 The solution set is Check Answer Seareh...
5 answers
6.1.27 Taoddct71nnea Ino Tirelo: trung?atalt Ehntomgy Tnande TnalniuDuon noasuiothrts pck#nnar (5kid47ije0hen Ouletmtennd une MclcumemeG0iz7ot: Cegra as ntede] )TtuloIRoundno4 eraltatort stbdlet Ind condd cikic" Momat4 [ fecoilty Taen [TYD Fan Laa ~eatn kutan J dertha monntng L doe nrul Ancin Ceecam #enmnenetelDniaoeoFanenellnt [n Wacoa(neou mnents tr teechuron wth U# Uesrale angkJulOneGaTeefmfteteln4htaualetaan uMTrchenEluinac (enatjuC,-D'
6.1.27 Taoddct 71nnea Ino Tirelo: trung?atalt Ehntomgy Tnande Tnalniu Duon noasuiothrts pck# nnar (5kid47ije 0hen Ouletmte nnd une MclcumemeG 0iz7ot: Cegra as ntede] ) Ttulo IRoundno4 eralta tort stbdlet Ind condd cikic" Momat4 [ fecoilty Taen [TYD Fan Laa ~eatn kutan J dertha monntng L doe nr...
2 answers
2. Verify that the Legendre Polynomials Po; P, Pz; and Pz satisfy Legendre $ equation [10 points]: (1~ r)dy - 2ady +n(n + l)y = 0 dx2 dx where (2n 2m) ! Pn(c) = E(-1)" n-2m m=0 2nm!(n m)l(n 2m)!
2. Verify that the Legendre Polynomials Po; P, Pz; and Pz satisfy Legendre $ equation [10 points]: (1~ r)dy - 2ady +n(n + l)y = 0 dx2 dx where (2n 2m) ! Pn(c) = E(-1)" n-2m m=0 2nm!(n m)l(n 2m)!...
5 answers
Four identical particles $A, B, C$ and $D$ move with equal speed $v$, along the diagonal of a square towards the centroid $O$ o ${$ the square, beginning at the vertices. $Lambda$ Gammaier the collision at $O, Lambda$ and $C$ retrace their path with specd $2 v$ and $B$ retraces its path with speed $v sqrt{2}$. Speed of the particle $D$ after the collision is(a) $sqrt{2} v$(b) $v$(c) $frac{v}{sqrt{2}}$(d) $2 v$
Four identical particles $A, B, C$ and $D$ move with equal speed $v$, along the diagonal of a square towards the centroid $O$ o ${$ the square, beginning at the vertices. $Lambda$ Gammaier the collision at $O, Lambda$ and $C$ retrace their path with specd $2 v$ and $B$ retraces its path with speed $...
5 answers
Choose the correct answer Find f' (x) if f (x)=cosx tanx Sil.xcOSXsinX(D) CUS.
Choose the correct answer Find f' (x) if f (x)=cosx tanx Sil.x cOSX sinX (D) CUS....
5 answers
@uebtiorpnoior with Powc 42 LW pulla upward ( load of tast 20 * 10' ke How [be ground Icvcl t0 lone - Would uke 200 m high foor at thc motor Fnae the [14tt Grot coualant pccd? 5 hout 0 Z9haat 0 Mhourt 70 4 hout 0 0 19 hout
@uebtior pnoior with Powc 42 LW pulla upward ( load of tast 20 * 10' ke How [be ground Icvcl t0 lone - Would uke 200 m high foor at thc motor Fnae the [14tt Grot coualant pccd? 5 hout 0 Z9haat 0 Mhourt 70 4 hout 0 0 19 hout...
5 answers
Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Then sketch the hyperbola using the asymptotes as an aid.$$ rac{(y+6)^{2}}{1 / 9}- rac{(x-2)^{2}}{1 / 4}=1$$
Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Then sketch the hyperbola using the asymptotes as an aid. $$\frac{(y+6)^{2}}{1 / 9}-\frac{(x-2)^{2}}{1 / 4}=1$$...
5 answers
Solve sin a tan asin a for all a € [0,2t)
Solve sin a tan a sin a for all a € [0,2t)...
1 answers
6. (24 pts) Write the triple intcgral; but DO NOT EVALUATE, by the graphs of z=4-r2 that represents the volume of the region bounded ~Y and z = 1- 2y region of It'$ not necessary to graph the solid, but it will help to graph the intersection itegthtion in the xY-planc. Hint: To find the region of integration in she xyoplanev findthe curvpbof of the two equations
6. (24 pts) Write the triple intcgral; but DO NOT EVALUATE, by the graphs of z=4-r2 that represents the volume of the region bounded ~Y and z = 1- 2y region of It'$ not necessary to graph the solid, but it will help to graph the intersection itegthtion in the xY-planc. Hint: To find the region ...
5 answers
Question25.0-V potential dillerence aCrOsS 25.00- W resistor What the curant flaiing through 1? You measure 4.00 A b. 5.00 A 1.00 d. 125 A
Question 25.0-V potential dillerence aCrOsS 25.00- W resistor What the curant flaiing through 1? You measure 4.00 A b. 5.00 A 1.00 d. 125 A...
5 answers
24 CH,OH Indentify the type of glycosidic 1 linkage the following carbohydrate:OH0 9 ? 6 6 W
24 CH,OH Indentify the type of glycosidic 1 linkage the following carbohydrate: OH 0 9 ? 6 6 W...
5 answers
Hjco(-:u) Which of the following reagents will not be useful in the synthesis of an epoxide _ CHz OsO4 (cat) (B) (C) (D) OOH OOH NaOHsingle step?
Hjco (-:u) Which of the following reagents will not be useful in the synthesis of an epoxide _ CHz OsO4 (cat) (B) (C) (D) OOH OOH NaOH single step?...
4 answers
Let A and Bbe two real symmetrc matrice s with the sameeigervaluecortesponding multiplicities are Ihc sare_ Prove that these an orthogonal invertble matx = such that PAPT _
Let A and Bbe two real symmetrc matrice s with the same eigervalue cortesponding multiplicities are Ihc sare_ Prove that these an orthogonal invertble matx = such that PAPT _...

-- 0.021016--