3

Graph appended to question

So I did:

$$\eqalign{ & y = {1 \over x^2} \cr & {x^2} = {1 \over y} \cr & x = y^{ - {1 \over 2}} \cr & \int_a^b x \, dy = \left[ 2{y^{1 \over 2}} \right]_{1 \over 4}^1 \cr & = 2\sqrt 1 - 2\sqrt {1 \over 4} \cr & = 2 - 1 \cr & = 1 \cr} $$

The answer however is 2, where have I gone wrong and misunderstood the question?

Thank you!

Michael Hardy
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seeker
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2 Answers2

6

You need to account for symmetry, because, recall, the square root function returns only the positive root. So you actually found the area under the curve strictly within the first quadrant. That is you found the area under only $x = +\frac{1}{\sqrt y}$

But actually, $x = \pm \frac{1}{\sqrt y}$, so we need $2 \times$ the area you computed. Using just $x = +\frac{1}{\sqrt y}$ is fine, but then double the area computed.

amWhy
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  • So say this wasn't a symmetrical cross section, how would I find the cross section of the other quadrant? – seeker May 12 '13 at 22:32
  • Here, that is irrelevant, as taking the sqrt of the function gives us $x = \pm \dfrac 1{\sqrt y}$, so computing the area under $-\dfrac 1{\sqrt y}$ to $+ \dfrac 1{\sqrt y}$ is the same as doubling the area under $\dfrac 1{\sqrt y}$. If you were to draw $x$ as a function of $y$, for example, exchange axes, with y the horizontal axis, and x the vertical axis, then you would the need for integrating between different bounds, or doubling the area that is "above" the $y$ axis. – amWhy May 12 '13 at 22:40
  • Also nice feedback and follow up! +1 – Amzoti May 13 '13 at 00:40
2

The cross-section is twice as large as you think it is. You have only calculated the first quadrant area. To put it another way, the distance between $x=-y^{-1/2}$ and $x=y^{-1/2}$ is $2y^{-1/2}$.

Remark: Sometimes, because we are so accustomed to integrating with respect to $x$, it can be useful to interchange the roles of $x$ and $y$. If you do that, and draw the appropriate picture, you are more likely to get things right.

André Nicolas
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