How can the area of a region bounded by a graph be found when given an integral?

[steelphantom]

This problem has me stumped because I can't figure out how to find the area of the region. I got an integral, but I don't know how to evaluate it. Here's the problem:

Find the area of the region bounded by the graph of $$y^2 = x^2 - x^4$$.

I solved the equation for y and got $$y = \pm\sqrt{x^2 - x^4}$$.

If I graph it, it looks like a bow tie with four symmetrical regions, so I decided to find the area of the top-right region. Here's the integral I came up with:

$$\int_{0}^{1} \sqrt{x^2 - x^4} dx$$.

That's all well and good I guess, but I have no idea how to evaluate that integral! As far as I know, I don't yet have the calculus knowledge to solve that type of integral. Should I solve for x and put the integral in terms of y?

 

[HallsofIvy]

This problem has me stumped because I can't figure out how to find the area of the region. I got an integral, but I don't know how to evaluate it. Here's the problem:

Find the area of the region bounded by the graph of $$y^2 = x^2 - x^4$$.

I solved the equation for y and got $$y = \pm\sqrt{x^2 - x^4}$$.

If I graph it, it looks like a bow tie with four symmetrical regions, so I decided to find the area of the top-right region. Here's the integral I came up with:

$$\int_{0}^{1} \sqrt{x^2 - x^4} dx$$.

That's all well and good I guess, but I have no idea how to evaluate that integral! As far as I know, I don't yet have the calculus knowledge to solve that type of integral. Should I solve for x and put the integral in terms of y?

Did you consider factoring:
$$\sqrt{x^2- x^4}= \sqrt{x^2(1- x^2)}= x\sqrt{1- x^2}$$
Looks like an easy substitution now.

 

[steelphantom]

Duh! That was it! I did that and got an area of 2/3 for one segment, so I multiplied by 4 and got a total area of 8/3 for the whole thing. Does that sound right? Thanks for the help!

 

[HallsofIvy]

Duh! That was it! I did that and got an area of 2/3 for one segment, so I multiplied by 4 and got a total area of 8/3 for the whole thing. Does that sound right? Thanks for the help!

Did you take into account the factor of 1/2 in $-\frac{1}{2}du= xdx$?

 

[steelphantom]

Did you take into account the factor of 1/2 in $-\frac{1}{2}du= xdx$?

Sorry about the delayed response; I was gone for a few days. I kind of rushed through it and must have skipped the factor. So the area would be 4/3 then, correct?

 

[HallsofIvy]

$$\int_{0}^{1} \sqrt{x^2 - x^4} dx= \int_0^1 \sqrt{1- x^2}x dx$$

Let $u= 1- x^2$ so $du= -2x dx$ or $-\frac{1}{2}du= x dx$. When x= 0, u= 1 and when x= 1, u= 0 so the integral becomes
$$-\frac{1}{2}\int_1^0 u^{\frac{1}{2}}du= \int_0^1 u^{\frac{1}{2}}du= \frac{1}{2}\fra{2}{3}u^{\frac{3}{2}\right}_{u=0}^1= \frac{1}{3}$$
Each segment has area 1/3 so all 4 have area 4/3.