Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Merciless
Group Title
Let R be the planar region between the curves y^2 + (x1)^2 = 4 and y^2 = 3(x1) which contains the point (0, 0). 1.) Calculate the total area of R. 2.) Determine the total length of the boundary of R.
 one year ago
 one year ago
Merciless Group Title
Let R be the planar region between the curves y^2 + (x1)^2 = 4 and y^2 = 3(x1) which contains the point (0, 0). 1.) Calculate the total area of R. 2.) Determine the total length of the boundary of R.
 one year ago
 one year ago

This Question is Closed

Merciless Group TitleBest ResponseYou've already chosen the best response.0
Here's the plot, x going from 1 to 3, y from (squareroot(3)) to squareroot(3)
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
dw:1354694787271:dw
 one year ago

Merciless Group TitleBest ResponseYou've already chosen the best response.0
??? Isn't R the region in between?
 one year ago

Merciless Group TitleBest ResponseYou've already chosen the best response.0
dw:1354694904341:dw
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
dw:1354694942463:dw
 one year ago

Merciless Group TitleBest ResponseYou've already chosen the best response.0
I still have no idea how to calculate the total area or boundary
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
so the region is the area of a circle , minus that parabola bit
 one year ago

Merciless Group TitleBest ResponseYou've already chosen the best response.0
makes sense
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
can you find the radius of the circle
 one year ago

Merciless Group TitleBest ResponseYou've already chosen the best response.0
the radius is 2
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
so the area of the circle is ...
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
right
 one year ago

Merciless Group TitleBest ResponseYou've already chosen the best response.0
and then?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
maybe some integration
 one year ago

Merciless Group TitleBest ResponseYou've already chosen the best response.0
that's exactly what I need, would love to learn how
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
have you found the intersection points ?
 one year ago

Merciless Group TitleBest ResponseYou've already chosen the best response.0
intersections at (2, squareroot(3)), and (2, squareroot(3))
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
dw:1354696097101:dw
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
lets calculate this area dw:1354696370978:dw
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
we can find the area of this region by integrating dw:1354696620853:dw \[=\int\limits_{x=1}^{x=2}\int\limits_{y=0}^{y=\sqrt{3(x1)}}\text dy\text dx\] \[=\int\limits_{y=0}^{y=\sqrt{3}}\int\limits_{x=\tfrac{y^2}3+1 }^{x=2}\text dx\text dy\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
maybe there is any easier way im not seeing
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
youre doing ok, you just seem to be cutting it up way to much
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
y^2 + (x1)^2 = 4 and y^2 = 3(x1) move it so that the circle is centered at the origin; the area stay the same, we just move it y^2 + (x1+1)^2 = 4 and y^2 = 3(x1+1) y^2 + x^2 = 4 and y^2 = 3x x = sqrt(2^2y^2) and x = 1/3 y^2 and it really doesnt matter if we have this oreiented long the x or y soo if its simpler to play with as y = sqrt(2^2x^2) and y = 1/3 x^2 thats fine too
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
so, what does all this moving and rewrite do? dw:1354733679539:dw it just makes it easier to see the equation to integrate with
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
the area to remove is between the circle and parabola \[\int_{a}^{b}\sqrt{4x^2}\frac13x^2~dx\]
 one year ago

Merciless Group TitleBest ResponseYou've already chosen the best response.0
and what is a,b?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
the intersections
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
+ sqrt(3)
 one year ago

Merciless Group TitleBest ResponseYou've already chosen the best response.0
Thanks! What about the total length of the boundary of R?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
hardest part might be in determining the length of the parabolas arc, circles are a dime a dozen
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
do you recall any details about how to work a line integral?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
dw:1354757262089:dw
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
\[\int_{C}ds=\int_{a}^{b}\sqrt{(x')^2+(y' )^2}\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
since x = x x' = 1 y = 1/3 x^2 y' = 2/3 x \[\int_{a}^{b}\sqrt{(1)^2+(\frac23x )^2}~dx\]
 one year ago

Merciless Group TitleBest ResponseYou've already chosen the best response.0
that integral equals the boundary?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
the integral equals the length of the curve of the parabola im sure you can figure out the circle part
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
dw:1354758192335:dw
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
find that and times 2 for the circle part
 one year ago

Merciless Group TitleBest ResponseYou've already chosen the best response.0
dw:1354758715223:dw
 one year ago

Merciless Group TitleBest ResponseYou've already chosen the best response.0
so 2x 1 + the integral of the curve (4.4) and I'd have 6.4 as total boundary length?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
\[y=\sqrt{4x^2}\] \[y'=\frac{x}{\sqrt{4x^2}}\] \[\int_{2}^{\sqrt3}\sqrt{1+\frac{x^2}{4x^2}}dx\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
circle part is, already did the parabola \[2\int_{2}^{\sqrt3}\sqrt{1+\frac{x^2}{4x^2}}dx\] \[2\int_{2}^{\sqrt3}\sqrt{\frac{4}{4x^2}}dx\] \[2\int_{2}^{\sqrt3}\frac{2}{\sqrt{{4x^2}}}dx\] \[dx=2sin(u)~:~dx=2cos(u) du\] \[2\int_{2}^{\sqrt3}\frac{2}{\sqrt{{44sin^2(u)}}}2cos(u) du\] \[2\int_{2}^{\sqrt3}\frac{2}{2\sqrt{{1sin^2(u)}}}2cos(u) du\] \[2\int_{2}^{\sqrt3}\frac{2}{2cos^2(u)}2cos(u) du\] \[2\int_{2}^{\sqrt3}\frac{2}{cos(u)} du\] \[4\int_{2}^{\sqrt3}sec(u) du\] \[4(ln(sec(u)+tan(u))\]and either adjust back to xs, or adujst the limits in terms of us
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.