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Merciless
Let R be the planar region between the curves y^2 + (x-1)^2 = 4 and y^2 = 3(x-1) which contains the point (0, 0). 1.) Calculate the total area of R. 2.) Determine the total length of the boundary of R.
Here's the plot, x going from -1 to 3, y from -(squareroot(3)) to squareroot(3)
|dw:1354694787271:dw|
??? Isn't R the region in between?
|dw:1354694904341:dw|
|dw:1354694942463:dw|
I still have no idea how to calculate the total area or boundary
so the region is the area of a circle , minus that parabola bit
can you find the radius of the circle
so the area of the circle is ...
maybe some integration
that's exactly what I need, would love to learn how
have you found the intersection points ?
intersections at (2, squareroot(3)), and (2, -squareroot(3))
|dw:1354696097101:dw|
lets calculate this area |dw:1354696370978:dw|
we can find the area of this region by integrating |dw:1354696620853:dw| \[=\int\limits_{x=1}^{x=2}\int\limits_{y=0}^{y=\sqrt{3(x-1)}}\text dy\text dx\] \[=\int\limits_{y=0}^{y=\sqrt{3}}\int\limits_{x=\tfrac{y^2}3+1 }^{x=2}\text dx\text dy\]
maybe there is any easier way im not seeing
youre doing ok, you just seem to be cutting it up way to much
y^2 + (x-1)^2 = 4 and y^2 = 3(x-1) move it so that the circle is centered at the origin; the area stay the same, we just move it y^2 + (x-1+1)^2 = 4 and y^2 = 3(x-1+1) y^2 + x^2 = 4 and y^2 = 3x x = sqrt(2^2-y^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(2^2-x^2) and y = 1/3 x^2 thats fine too
so, what does all this moving and rewrite do? |dw:1354733679539:dw| it just makes it easier to see the equation to integrate with
the area to remove is between the circle and parabola \[\int_{a}^{b}\sqrt{4-x^2}-\frac13x^2~dx\]
Thanks! What about the total length of the boundary of R?
hardest part might be in determining the length of the parabolas arc, circles are a dime a dozen
do you recall any details about how to work a line integral?
|dw:1354757262089:dw|
\[\int_{C}ds=\int_{a}^{b}\sqrt{(x')^2+(y' )^2}\]
since x = x x' = 1 y = 1/3 x^2 y' = 2/3 x \[\int_{a}^{b}\sqrt{(1)^2+(\frac23x )^2}~dx\]
that integral equals the boundary?
the integral equals the length of the curve of the parabola im sure you can figure out the circle part
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find that and times 2 for the circle part
|dw:1354758715223:dw|
so 2x 1 + the integral of the curve (4.4) and I'd have 6.4 as total boundary length?
\[y=\sqrt{4-x^2}\] \[y'=\frac{-x}{\sqrt{4-x^2}}\] \[\int_{-2}^{\sqrt3}\sqrt{1+\frac{x^2}{4-x^2}}dx\]
circle part is, already did the parabola \[2\int_{-2}^{\sqrt3}\sqrt{1+\frac{x^2}{4-x^2}}dx\] \[2\int_{-2}^{\sqrt3}\sqrt{\frac{4}{4-x^2}}dx\] \[2\int_{-2}^{\sqrt3}\frac{2}{\sqrt{{4-x^2}}}dx\] \[dx=2sin(u)~:~dx=2cos(u) du\] \[2\int_{-2}^{\sqrt3}\frac{2}{\sqrt{{4-4sin^2(u)}}}2cos(u) du\] \[2\int_{-2}^{\sqrt3}\frac{2}{2\sqrt{{1-sin^2(u)}}}2cos(u) du\] \[2\int_{-2}^{\sqrt3}\frac{2}{2cos^2(u)}2cos(u) du\] \[2\int_{-2}^{\sqrt3}\frac{2}{cos(u)} du\] \[4\int_{-2}^{\sqrt3}sec(u) du\] \[4(ln(sec(u)+tan(u))\]and either adjust back to xs, or adujst the limits in terms of us