Finding the area bounded by the curves

[shamieh]

So it's been a while since I've done one of these problems. Need to make sure I am using the right procedures to solve it.

Q)Find the area bounded by the curve $y = \frac{1}{2}x^2$ and $x^2 + y^2 = 8$

So first thing I did was plug in numbers to get the two graphs. It looks like they intersect at the point (3,3), I'm not sure if I'm right or wrong on that.

Don't i need to set the two equations equal to each other to find the points of intersection?

So would I be correct in saying: $\frac{1}{2}x^2 = x^2 + y^2 - 8$?

Proceeding through I get:
$x^2 = 2x^2 + 2y^2 - 8$
$=x^2 + 2y^2 - 16$ <-- I know this is simple but I'm stuck because the two different variables are throwing me off (x & y)

Any advice or guidance would be great. Thanks in advance.

 

[Ackbach]

I'm afraid you need to improve your method of finding the intersection points. Setting $x^{2}/2=x^{2}+y^{2}-8$ is comparing apples and oranges. Instead, you need to set $y_{1}=y_{2}$. That is, the $y$ coordinate from one graph must equal the $y$ coordinate from the other graph. So, you should set $x^{2}/2= \pm \sqrt{8-x^{2}}$. Or, a slightly more clever way would be to eliminate $x^{2}$: $2y=x^{2}$, so $2y+y^{2}=8$. In any case, double-check any coordinates you get against your original functions. $(3,3)$, e.g., is not on either of your curves!

 

[shamieh]

I see. So now I have $\frac{x^2}{2} = +- \sqrt{8x - 2}$ So now I just solve correct?

so I need to do this? $(\frac{x^2}{2})^2 = (\sqrt{8 - x^2})^2$

Because now I end up with this ugly thing: $\frac{x^4}{4} = 8 - x^2$ and what should I do with that? multiply by $\frac{4}{1}$ to get rid of the fraction or what? I'm a bit confused. I noticed you eliminated $x^2$ easily but I'm lost on how exactly you did that to get $2y + y^2 - 8$.

 

[Ackbach]

I see. So now I have $\frac{x^2}{2} = +- \sqrt{8x - 2}$ So now I just solve correct?

so I need to do this? $(\frac{x^2}{2})^2 = (\sqrt{8 - x^2})^2$

Because now I end up with this ugly thing: $\frac{x^4}{4} = 8 - x^2$ and what should I do with that?

This is a quadratic in $x^{2}$, so solve for $x^{2}$ first using the quadratic formula, and then solve each of those (if they are positive!) for $x$.

multiply by $\frac{4}{1}$ to get rid of the fraction or what? I'm a bit confused. I noticed you eliminated $x^2$ easily but I'm lost on how exactly you did that to get $2y + y^2 - 8$.

You have two equations:
\begin{align*}
y&= \frac{1}{2} x^{2} \\
8&=x^{2}+y^{2}.
\end{align*}
Take the first equation and solve for $x^{2}$ to obtain $2y=x^{2}$. Then, in the second equation, everywhere you see an $x^{2}$, replace with $2y$. You get
\begin{align*}
8&=2y+y^{2} \\
y^{2}+2y-8&=0.
\end{align*}
Then solve this quadratic.

 

[shamieh]

So let me see if I understand what you are saying correctly.

I need to set up as $y_1 = y_2$

So I have $y = \frac{1}{2}x^2$ and $x^2 + y^2 = 8$

So I can easily multiply by the reciprocal and get $2y = x^2$

Thus: $2y + y^2 = 8 = y^2 + 2y -8 = (y - 2)(y + 4)$

So then would my integral be from $\int^{-4}_{2}$ and won't that be a problem?

 

[Ackbach]

So let me see if I understand what you are saying correctly.

I need to set up as $y_1 = y_2$

So I have $y = \frac{1}{2}x^2$ and $x^2 + y^2 = 8$

So I can easily multiply by the reciprocal and get $2y = x^2$

Thus: $2y + y^2 = 8 = y^2 + 2y -8 = (y - 2)(y + 4)$

So then would my integral be from $\int^{-4}_{2}$ and won't that be a problem?

I think you mean $ \displaystyle \int_{-4}^{2} g(y) \, dy$. That would be if it makes more sense for you to divvy up the $y$ interval. Can you draw the region enclosed?

 

[shamieh]

Wait? I thought that $y^2 + 2y - 8 = (y - 2)(y + 4)$ THUS $y = 2, y = - 4$ So do you just make the integral starting from the lowest number to the largest number or do you have to put them on the integral in the order you solved them? For example since $y - 2 = 2$, and that came first would that mean I would be starting from $2$ and ending at a $-4$? OR are you just saying that it is from $\int^{-4}_2$ and you just put a $(-)$ in front of the integral to change it to $\int^2_{-4}$

 

[Ackbach]

Wait? I thought that $y^2 + 2y - 8 = (y - 2)(y + 4)$ THUS $y = 2, y = - 4$ So do you just make the integral starting from the lowest number to the largest number or do you have to put them on the integral in the order you solved them? For example since $y - 2 = 2$, and that came first would that mean I would be starting from $2$ and ending at a $-4$? OR are you just saying that it is from $\int^{-4}_2$ and you just put a $(-)$ in front of the integral to change it to $\int^2_{-4}$

Generally, you put the values from lowest to highest. However, the bigger question right now is whether you want to integrate w.r.t. $x$ or $y$. For each of the $y$ values you found, you need to find the corresponding $x$ values. You also need to draw the region so you can decide how you want to integrate.

 

[shamieh]

Right. So I will have horizontal asymptotes both at $-4$ and $2$. So I want to say top function minus the bottom function. So $\int^2_{-4} (2y) - (y^2 - 8) \,dy$

 

[Ackbach]

Right. So I will have horizontal asymptotes both at $-4$ and $2$. So I want to say top function minus the bottom function. So $\int^2_{-4} (2y) - (y^2 - 8) \,dy$

I'm afraid not. You need to do two things before you can possibly continue correctly:

1. You need to find the $x$-coordinates corresponding to the $y$ coordinates for which you solved.
2. You need to DRAW THE REGION!

 

[shamieh]

so, would I have $\int^2_{-2} (\sqrt{8 - x^2}) - (\frac{x^2}{2}) \, dx$ ?

And if I split those up into $TWO$ different integrals. I did u substitution for the first one, $\sqrt{8-x^2}$ and I'm ending up with 0 once I update the limits...Is that incorrect?

 

[Ackbach]

You finally do have a correct integral. $u$-sub won't work on the first integral, because you don't have a derivative of any inside function present. I would go with a trig substitution. The $\displaystyle \int_{-2}^{2}(x^{2}/2) \, dx$ integral should be straight-forward.