MHB Lazi's question at Yahoo Answers regarding the area bounded by two functions

AI Thread Summary
The discussion focuses on calculating the area between the curves defined by the equations x=y^2 and x=4-y^2. The area is determined by shifting the functions and using symmetry to simplify the integration process. The final calculated area is confirmed to be (16√2)/3. A detailed solution is provided, illustrating the integration steps involved. This method effectively demonstrates how to find the area between the specified curves.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

Find the area between those curves: x=y^2, and x=4-y^2?

Find the area between those curves: x=y^2, and x=4-y^2

the answer should be (16 sqrt2 ) /3

Here is a link to the question:

Find the area between those curves: x=y^2, and x=4-y^2? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
Mathematics news on Phys.org
Hello Lazi,

I would orient the coordinate axes such that the $x$-axis is vertical, and the $y$-axis is horizontal. To further simplify matters, I would vertical shift both functions down two units, then use symmetry so that we wish to find 4 times the following shaded area:

2hckkl2.jpg


Hence the desired area $A$ is:

$$A=4\int_0^{\sqrt{2}}2-y^2\,dy=\frac{4}{3}\left[6y-y^3 \right]_0^{\sqrt{2}}=\frac{4}{3}\left(6\sqrt{2}-2\sqrt{2} \right)=\frac{16\sqrt{2}}{3}$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top