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I am given $y=x$ and $y=\frac{x}{8}$ for the lines. The curve is $y=\frac{1}{x^2}$ . This is more of a question about the question though. I have no idea what it is asking.

My instinct is to integrate over some limits but I am not sure where this circle features and whether I am over complicating things.

Should I just be adding 3 indefinite integrals here?

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You write "where this circle features", but you don't mention a circle anywhere else. – joriki Aug 12 '12 at 10:29
Hint: The complement of the curves $y=x$, $y=x/8$ and $y=1/x^2$ in the plane is made of 8 connected components (if I am not mistaken), only one of them being bounded. The question is to compute the area of this bounded region and this type of homework is designed to make you understand the initial steps I described, and in particular, to determine the bounded connected component. – Did Aug 12 '12 at 10:34
@joriki ha you're right. I should not have assumed that. – Magpie Aug 12 '12 at 10:56
Should not have assumed what? – joriki Aug 12 '12 at 10:57
that the curve was a circle. I don't know why I did that I have it as an asymptote! – Magpie Aug 12 '12 at 11:06
up vote 1 down vote accepted

Make a graph showing all three curves. You’ll see that there is just one finite region bounded by these curves. It has two straight sides along the lines $y=x$ and $y=x/8$ and one curved side along $y=1/x^2$. If that third side ran straight from $\langle 1,1\rangle$ to $\left\langle 2,\frac14\right\rangle$, the region would be the triangle with vertices at the origin, at $\langle 1,1\rangle$, and at $\left\langle 2,\frac14\right\rangle$.

(In case it isn’t clear, $\langle 1,1\rangle$ is the point of intersection of $y=x$ and $y=1/x^2$, and $\left\langle 2,\frac14\right\rangle$ is the point of intersection of $y=x/8$ and $y=1/x^2$.)

You’ll want to set it up as two integrals, one from $x=0$ to $x=1$, the other from $x=1$ to $x=2$. Your vertical strips $dA$ of area for the first integral will run between the two straight lines; for the second integral they’ll run from the line $y=x/8$ to the curve $y=1/x^2$.

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Thanks, that makes sense. The graph really helped I managed to find the points after that, but I wasn't sure what to do with them them until I checked this again! – Magpie Aug 12 '12 at 11:05
ok so I have got $$\int_0^{1}xdx+\int_1^{2}\frac{1}{x^2}dx-\frac{1}{8}\int_1^{2}xdx$$ my latex is no good. Yay! – Magpie Aug 12 '12 at 11:41
if anyone can confirm whether or not i got this step ok. That would be grand! – Magpie Aug 12 '12 at 11:46
@Magpie: Yes, that looks fine. – Brian M. Scott Aug 12 '12 at 11:56

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