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 one year ago
Ok the attached image is from a calculus textbook (http://www.stewartcalculus.com/media/11_home.php).
The question is simple enough  it says that this figure contains all points inside the square that are closer to the center than to the sides of the square. Find the area of this region.
Anyone have an idea on how to get started?
The image will be uploaded shortly.
 one year ago
Ok the attached image is from a calculus textbook (http://www.stewartcalculus.com/media/11_home.php). The question is simple enough  it says that this figure contains all points inside the square that are closer to the center than to the sides of the square. Find the area of this region. Anyone have an idea on how to get started? The image will be uploaded shortly.

This Question is Closed

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.2I think, you'll need to do 2 integrals, and then using symmetry you should be able to apply these to the rest of the diagram. However, I'm not sure of the best way to set up the integrals. dw:1356844485378:dw This picture very roughly imitates your shape. The areas I've shaded need to be integrated, each one separately. The area on the left you can find by looking at the center of the square and the left side of the box. The area in the middle can be found by looking at the center of the square and the upper part of the square. You just need to find the equation of the line that is equidistant from the corresponding sides of the square and the center of the square. If I remember correctly, these should take the form of hyperbolas.

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.2Alright, I've got to go. Hopefully this was enough to get you started. I'm pretty sure that the lines you're looking for will be hyperbolic. Then you just have a couple definite integrals to solve.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for your time!

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0I will work on this right away

wio
 one year ago
Best ResponseYou've already chosen the best response.1It also might be interesting to note that a parabola is defined as a curve where all points are equidistant to a focus (point) and a directrix (line). so you can tell that these lines are parabolas.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0Hm. I was thinking parabolas, too. I think that might be just what I am looking for.

wio
 one year ago
Best ResponseYou've already chosen the best response.1From my calculus book: An equation of a (horizontal) parabola with focus \((0, p)\) and directrix \(y = p\) is: \(x^2 = 4py\).

wio
 one year ago
Best ResponseYou've already chosen the best response.1Let the very bottom line be the directrix and let the very center be the focus...

wio
 one year ago
Best ResponseYou've already chosen the best response.1The center and directrix are 1 unit apart, so \[ p(p) = 1 \\ 2p = 1 \\ p = 1/2 \] Obviously \(p\). can't be negative

wio
 one year ago
Best ResponseYou've already chosen the best response.1\[ x^2 = 4(1/2)y = 2y \implies y = x^2/2 \]So this gives us the equation of the bottom line when the xaxis is the very center, and the y axis is halfway between the center (focus) and the bottom (directrix). @LogicalApple I think you can do the rest from here.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much!

wio
 one year ago
Best ResponseYou've already chosen the best response.1No problem, it was a bit of a learning experience for me as well.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0Would you agree with the points I drew in and labeled ?

wio
 one year ago
Best ResponseYou've already chosen the best response.1Actually, let me think about it for a moment...

wio
 one year ago
Best ResponseYou've already chosen the best response.1I'm not sure of the logic used to get those points so I can't really confirm or deny... I am thinking perhaps you could find where the two parabolas would intercept.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0I based my calculations on these points. I had enough points to generate parabolas. One parabola was x = (2  sqrt 2)y^2  1 This would he the parabola on the far left side. Since I'm only interested in a portion of it (namely from when x = 1 to when x = sqrt (2) / 2) I could integrate y = sqrt( (1 + x) / (2  sqrt (2)) from 1 to sqrt (2)/2 Multiply this result by 4 and that would take care of the 4 triangular regions (I only drew two of them in the figure). For the parabola hanging overhead from x = sqrt(2)/2 to sqrt(2)/2 I calculated y = (sqrt 2  2)x^2 + 1. Integrate this from sqrt(2)/2 to sqrt(2)/2 and multiply the result by 2. This takes care of the remainder of the region I think. My end result was ( 8 sqrt(2)  2)/3 But I have no way of verifying this

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0The boundary of this figure is equally as far from the center as it is from the perimeter of the square. So along the horizontal axis, the point (1, 0) is as far from (0, 0) as it is from (2, 0). I suppose I could have used (0, 1) as another point.... Instead I determined the point of the corner of the boundary as half of the diagonal of a square of hypotenuse 4 (2^2 + 2^2).

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0Wait now that I think about it..

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0Those numbers should be cut in half

wio
 one year ago
Best ResponseYou've already chosen the best response.1dw:1356849810908:dw I know it's not a circle but think of it this way.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0You're absolutely right. Back to the drawing board for me.

wio
 one year ago
Best ResponseYou've already chosen the best response.1In this case, it was just finding the right algebra.... lol

wio
 one year ago
Best ResponseYou've already chosen the best response.1\[ \sqrt{x^2 + x^2} + x = \sqrt{1^2+1^2} = \sqrt{2} \]

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0Right, sqrt(2) . Hm.

wio
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, well, just solve for \(x\)...

wio
 one year ago
Best ResponseYou've already chosen the best response.1\[ \sqrt{x^2+x^2}+x = \sqrt{2x^2}+x = \sqrt{2} \cdot x+x \]Thus \[ (1+\sqrt{2})x = \sqrt{2} \]@LogicalApple Make sense... convinced? Don't assume I'm a master, either. Look at what I did to check for mistakes.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0I'm still trying to figure out why x + y = sqrt(1^2 +1^2) instead of x + y = sqrt(2^2 + 2^2)?

wio
 one year ago
Best ResponseYou've already chosen the best response.1More stuff:\[ (\sqrt{2}+1)x = \sqrt{2} \\ (\sqrt{2}1)(\sqrt{2}+1)x=(\sqrt{2}1)\sqrt{2} \\ (21)x = 2\sqrt{2} \\ x = 2\sqrt{2} \]

wio
 one year ago
Best ResponseYou've already chosen the best response.1Because the total length is 2, but we're only doing halfway... which is 1.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0Yeah scratch that

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0I am in agreement that x = 2  sqrt 2

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0I will modify my equations

wio
 one year ago
Best ResponseYou've already chosen the best response.1Okay, so what we can do if first just find the area of the inner square. Then we can find the area between one of the parabola and the inner square... and by symmetry just multiply that by 4.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0I like that idea a lot actually

wio
 one year ago
Best ResponseYou've already chosen the best response.1The area of the square is easy. We just found half the length of its side. So just double that and square it.

wio
 one year ago
Best ResponseYou've already chosen the best response.1We also already found the equation of the parabola, and we know its limits.

wio
 one year ago
Best ResponseYou've already chosen the best response.1@LogicalApple Got an idea for finding the area under that parabola?

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0I'm still uncertain of the equation of the parabola. I calculate something like: y = (sqrt 2  1) / (4 sqrt 2  6) * x^2 + 1

wio
 one year ago
Best ResponseYou've already chosen the best response.1I don't know how you're getting that...

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0This parabola contains the points (2 + sqrt 2, 2  sqrt 2), (0, 1), and (2  sqrt 2, 2  sqrt 2) The top two corner points of the square and the vertex of the parabola

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0I use a system of 3 equations. Where the general form is f(x) = ax^2 + bx + c I let x = 2 + sqrt 2, f(x) = 2  sqrt 2 x = 0, f(x) = 1 x = 2  sqrt 2, f(x) = 2  sqrt (2)

wio
 one year ago
Best ResponseYou've already chosen the best response.1We could just ignore that it a parabola altogether and just try using the distance formula.

wio
 one year ago
Best ResponseYou've already chosen the best response.1Distance of some point \((x, y)\) to the line \(y=1\) \[ \sqrt{(xx)^2(y1)^2} \]Distance of some point \((x, y)\) to the origin \[ \sqrt{x^2+y^2} \]Set them equal \[ \sqrt{(xx)^2(y1)^2} = \sqrt{x^2+y^2} \]

wio
 one year ago
Best ResponseYou've already chosen the best response.1Should be a plus in there.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0Now I think I'm getting a rational answer of 8/3 for the area of the region.

wio
 one year ago
Best ResponseYou've already chosen the best response.1\[ \sqrt{(y1)^2} = \sqrt{x^2 + y^2} \\ (y1)^2 = x^2 + y^2 \\ y^2  2y + 1 = x^2 + y^2 \\ 2y + 1 = x^2 \\ y = \frac{x^2}{2} +\frac{1}{2} \]We got a different equation because we had a different origin.

wio
 one year ago
Best ResponseYou've already chosen the best response.1But notice how our slope is pretty much the same.

wio
 one year ago
Best ResponseYou've already chosen the best response.1Just reflected because this time we took the concave down parabola at top.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0I'm following.. and I'm curious. I am going to review this entire post when it's all said and done.

wio
 one year ago
Best ResponseYou've already chosen the best response.1To find the area under parabola \(y = x^2/2 + 1/2\) and above the line \(y = 2  \sqrt{2} \), we take the integral: \[ \int_{1(2\sqrt{2})}^{2\sqrt{2}} \left[ \frac{x^2 + 1}{2}  (2\sqrt{2}) \right]dx \]

wio
 one year ago
Best ResponseYou've already chosen the best response.1\[ \int_{(2\sqrt{2})}^{2\sqrt{2}} \left[ \frac{x^2 + 1}{2}  (2\sqrt{2}) \right]dx \]

wio
 one year ago
Best ResponseYou've already chosen the best response.1Sorry, been having quite a few typos today.

wio
 one year ago
Best ResponseYou've already chosen the best response.1That's a pretty easy integral... but it just has a lot of sqrts and stuff that I am a bit reluctant to do it.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0Right but I can just use symmetry again. Plus there is wolfram alpha

wio
 one year ago
Best ResponseYou've already chosen the best response.1No it's an extremely easy integral it's just:\[ \int_{(2\sqrt{2})}^{2\sqrt{2}} \left[ \frac{x^2 + 1}{2}  (2\sqrt{2}) \right]dx \\ = \left[ \frac{x^3}{3\times 2} + \frac{x}{2}  (2\sqrt{2})x \right] _{(2\sqrt{2})}^{2\sqrt{2}} \]

wio
 one year ago
Best ResponseYou've already chosen the best response.1plugging that pellet in that is gonna be a pain in the retrice..

wio
 one year ago
Best ResponseYou've already chosen the best response.1Anyway, I know you can do the rest from here. Just look things over a couple times if you don't get anything I said. There maybe some typos but the general methodology is correct as far as I can see.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{(2  \sqrt{2)}}^{2  \sqrt{2}} \frac{ \sqrt{2}  1 }{ 4\sqrt{2}  3} x^{2}dx \int\limits_{(2  \sqrt{2)}}^{2  \sqrt{2}} (2  \sqrt{2}) dx = 4\sqrt{2}  16\] The above integral is based on the attached picture (below) It determines the area of one of the parabolic regions bounded by the inner square. Multiply this result by 4 and add the area of the square, we obtain 8/3 as the area of the region.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0should be a dx in that second integral (sorry am still getting used to the equation creator)

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0Oh wait there is already one there. Nvm. Anyway, thanks a lot to @wio for your patience. This was an unusually challenging problem but you found a way to simplify the heck out of it! And thanks to KingGeorge for visualizing it and getting me on the right track. OpenStudy is pretty cool..

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0after looking this over...i agree with the methods except i believe the intersection points are incorrect the limits should be +(sqrt2 1) Not +(2sqrt2) when \[\frac{1}{2}x^{2}+\frac{1}{2} = x\] \[x = \sqrt{2}  1\]

wio
 one year ago
Best ResponseYou've already chosen the best response.1That looks right... makes me wonder where I went wrong.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0I think your geometry and algebra are correct in concluding the corner points of the region do lay at + (2  sqrt 2). I will double check.

wio
 one year ago
Best ResponseYou've already chosen the best response.1@LogicalApple They are somehow wrong. @dumbcow 's method makes more sense.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0But didn't you calculate the distance from a point to the line y = 1 instead of to the line y = 2?

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0Of course you did. Hm.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0Let me see what I obtain next.

wio
 one year ago
Best ResponseYou've already chosen the best response.1Wow it's really bothering me how I messed up. I can't figure out why my trig method was wrong.

wio
 one year ago
Best ResponseYou've already chosen the best response.1@dumbcow 's method makes sense... you want the point where our parabola intersects y = x... But my method seemed correct as well... gotta figure out the fallacy I made.

wio
 one year ago
Best ResponseYou've already chosen the best response.1Ooooh! I figured out my error!

wio
 one year ago
Best ResponseYou've already chosen the best response.1dw:1356855949526:dw We found \(x\)! We wanted \(z\) in the diagram!

wio
 one year ago
Best ResponseYou've already chosen the best response.1\[ 2z^2 = x^2 \\ \sqrt{2}\cdot z= x \\ z = x/\sqrt{2} = \frac{2\sqrt{2}}{\sqrt{2}} = \frac{2}{\sqrt{2}}  1 = \sqrt{2}1 \]Lol... really worry about that @LogicalApple Thank @dumbcow for spotting it.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0Modifying .. 1 sec

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0I did worry about it!

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0I will integrate now.. and I have a feeling it will be irrational this time ;(

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0I now see why there was a discrepancy with my earlier parabolic equation.

wio
 one year ago
Best ResponseYou've already chosen the best response.1I knew from the very start there was very little chance of a rational result.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0My image is still wrong. One more modification ..

wio
 one year ago
Best ResponseYou've already chosen the best response.1That's pretty much it. It's an easy integral because it's just a 2 degree polynomial.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0Final answer is ... \[\frac{ 4 }{ 3 }(4\sqrt{2}  5) \]

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0That is 4 times the parabola plus the area of the square.

wio
 one year ago
Best ResponseYou've already chosen the best response.1It lists your answer as an alternate form.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{(\sqrt{2}  1)}^{\sqrt{2} 1} (\frac{ 1  x ^{2} }{ 2}) dx  \int\limits_{(\sqrt{2}  1)}^{\sqrt{2} 1} (\sqrt{2}  1)dx = \frac{ 10 \sqrt{2} 14 }{ 3 }\] Oh yeah, Wolfram is great! The above should be the area of just one parabolic region. Multiplying this by 4 and then adding the area of the square yields the previous result.

wio
 one year ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=%5Cint+from+%28sqrt%282%29+1%29+to+%28sqrt%282%291%29+++%5B+%281x%5E2%29%2F2++%28sqrt%282%29++1%29%5Ddx Wolfram agreeing with that too. Good algebra.

LogicalApple
 one year ago
Best ResponseYou've already chosen the best response.0I feel like something was accomplished tonight. Then again, I wonder what the one guy felt like spending 7 years solving Fermat's last theorem....
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