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anonymous
 3 years ago
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.
anonymous
 3 years ago
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.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here's the plot, x going from 1 to 3, y from (squareroot(3)) to squareroot(3)

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1354694787271:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0??? Isn't R the region in between?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354694904341:dw

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1354694942463:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I still have no idea how to calculate the total area or boundary

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2so the region is the area of a circle , minus that parabola bit

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2can you find the radius of the circle

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2so the area of the circle is ...

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2maybe some integration

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's exactly what I need, would love to learn how

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2have you found the intersection points ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0intersections at (2, squareroot(3)), and (2, squareroot(3))

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1354696097101:dw

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2lets calculate this area dw:1354696370978:dw

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

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2maybe there is any easier way im not seeing

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1youre doing ok, you just seem to be cutting it up way to much

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1y^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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks! What about the total length of the boundary of R?

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1do you recall any details about how to work a line integral?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1354757262089:dw

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that integral equals the boundary?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1the integral equals the length of the curve of the parabola im sure you can figure out the circle part

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1354758192335:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1find that and times 2 for the circle part

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354758715223:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 2x 1 + the integral of the curve (4.4) and I'd have 6.4 as total boundary length?

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1circle 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
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