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anonymous
 3 years 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.
anonymous
 3 years 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
 3 years 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
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you for your time!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I will work on this right away

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It 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.

anonymous
 3 years 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.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let the very bottom line be the directrix and let the very center be the focus...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The center and directrix are 1 unit apart, so \[ p(p) = 1 \\ 2p = 1 \\ p = 1/2 \] Obviously \(p\). can't be negative

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ x^2 = 4(1/2)y = 2y \implies y = x^2/2 \]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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No problem, it was a bit of a learning experience for me as well.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Would you agree with the points I drew in and labeled ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Actually, let me think about it for a moment...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'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.

anonymous
 3 years 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

anonymous
 3 years 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).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wait now that I think about it..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Those numbers should be cut in half

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1356849810908:dw I know it's not a circle but think of it this way.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You're absolutely right. Back to the drawing board for me.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1356849937491:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1356850006505:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0In this case, it was just finding the right algebra.... lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \sqrt{x^2 + x^2} + x = \sqrt{1^2+1^2} = \sqrt{2} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, well, just solve for \(x\)...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \sqrt{x^2+x^2}+x = \sqrt{2x^2}+x = \sqrt{2} \cdot x+x \]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.

anonymous
 3 years 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)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0More stuff:\[ (\sqrt{2}+1)x = \sqrt{2} \\ (\sqrt{2}1)(\sqrt{2}+1)x=(\sqrt{2}1)\sqrt{2} \\ (21)x = 2\sqrt{2} \\ x = 2\sqrt{2} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Because the total length is 2, but we're only doing halfway... which is 1.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am in agreement that x = 2  sqrt 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I will modify my equations

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1356850703108:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I like that idea a lot actually

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The area of the square is easy. We just found half the length of its side. So just double that and square it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We also already found the equation of the parabola, and we know its limits.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@LogicalApple Got an idea for finding the area under that parabola?

anonymous
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't know how you're getting that...

anonymous
 3 years 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

anonymous
 3 years 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We could just ignore that it a parabola altogether and just try using the distance formula.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Distance 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 \[ \sqrt{(xx)^2(y1)^2} = \sqrt{x^2+y^2} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Should be a plus in there.

anonymous
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \sqrt{(y1)^2} = \sqrt{x^2 + y^2} \\ (y1)^2 = x^2 + y^2 \\ y^2  2y + 1 = x^2 + y^2 \\ 2y + 1 = x^2 \\ y = \frac{x^2}{2} +\frac{1}{2} \]We got a different equation because we had a different origin.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But notice how our slope is pretty much the same.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Just reflected because this time we took the concave down parabola at top.

anonymous
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To find the area under parabola \(y = x^2/2 + 1/2\) and above the line \(y = 2  \sqrt{2} \), we take the integral: \[ \int_{1(2\sqrt{2})}^{2\sqrt{2}} \left[ \frac{x^2 + 1}{2}  (2\sqrt{2}) \right]dx \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \int_{(2\sqrt{2})}^{2\sqrt{2}} \left[ \frac{x^2 + 1}{2}  (2\sqrt{2}) \right]dx \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry, been having quite a few typos today.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's a pretty easy integral... but it just has a lot of sqrts and stuff that I am a bit reluctant to do it.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No it's an extremely easy integral it's just:\[ \int_{(2\sqrt{2})}^{2\sqrt{2}} \left[ \frac{x^2 + 1}{2}  (2\sqrt{2}) \right]dx \\ = \left[ \frac{x^3}{3\times 2} + \frac{x}{2}  (2\sqrt{2})x \right] _{(2\sqrt{2})}^{2\sqrt{2}} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0plugging that pellet in that is gonna be a pain in the retrice..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Anyway, 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.

anonymous
 3 years 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.

anonymous
 3 years 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)

anonymous
 3 years 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
 3 years 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 \[\frac{1}{2}x^{2}+\frac{1}{2} = x\] \[x = \sqrt{2}  1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That looks right... makes me wonder where I went wrong.

anonymous
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@LogicalApple They are somehow wrong. @dumbcow 's method makes more sense.

anonymous
 3 years 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Of course you did. Hm.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let me see what I obtain next.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wow it's really bothering me how I messed up. I can't figure out why my trig method was wrong.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ooooh! I figured out my error!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1356855949526:dw We found \(x\)! We wanted \(z\) in the diagram!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1356856038380:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ 2z^2 = x^2 \\ \sqrt{2}\cdot z= x \\ z = x/\sqrt{2} = \frac{2\sqrt{2}}{\sqrt{2}} = \frac{2}{\sqrt{2}}  1 = \sqrt{2}1 \]Lol... really worry about that @LogicalApple Thank @dumbcow for spotting it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I did worry about it!

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I knew from the very start there was very little chance of a rational result.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0My image is still wrong. One more modification ..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's pretty much it. It's an easy integral because it's just a 2 degree polynomial.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For just the parabola?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That is 4 times the parabola plus the area of the square.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It lists your answer as an alternate form.

anonymous
 3 years 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.

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

anonymous
 3 years 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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