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We have an elliptic area defined by $$A := \{(x, y) \in \mathbb{R}^2 \mid (\tfrac{x}{a})^2+(\tfrac{y}{b})^2 \leq 1 \}$$ and a height function $$h \colon \mathbb{R}^2 \to \mathbb{R}, (x, y) \mapsto h(x, y) = c\left[1-(\tfrac{x}{a})^2-(\tfrac{y}{b})^2\right]$$ by transforming the area constraint we get $$y = \pm b\sqrt{1-(\tfrac{x}{a})^2}$$ which gives us the bounds for integrating over the y-axis. By setting $y = 0$ we get the bounds for the x-axis which is $\pm a$.

Finally we can write down our integral $$\int_Ah(x,y)dxdy = \int_{x_1=-a}^{x_2=a} \int_{y_1=-b\sqrt{1-(\tfrac{x}{a})^2}}^{y_2=b\sqrt{1-(\tfrac{x}{a})^2}}c\left[1-(\tfrac{x}{a})^2-(\tfrac{y}{b})^2\right]dydx = \\\ 4c\int_{0}^{x_2} \int_{0}^{y_2}\left[1-(\tfrac{x}{a})^2-(\tfrac{y}{b})^2\right]dydx$$ now comes the hard part.

1. Approach

This was my own attempt which does get stuck at some point

$$\dfrac{8}{3}cb\int_0^a\left[1-(\tfrac{x}{a})^2\right]^\frac{3}{2}dx \stackrel{x = a\sin(z)}{=} \dfrac{8}{3}\dfrac{cb}{a}\int_{z_1=0}^{z_2=a\sin(a)} \left[1-\sin(z)^2\right]^\frac{3}{2}\dfrac{1}{\cos(z)}dz = \dfrac{8}{3}\dfrac{cb}{a}\int_{z_1}^{z_2} \cos(z)^2dz = \dfrac{8}{3}\dfrac{cb}{a}\left[z+\cos(z)\sin(z)\right]\vert^{z_2}_{z_1} = \dfrac{8}{3}\dfrac{cb}{a}\left[a\sin(a)+\cos(a\sin(a))\sin(a\sin(a))\right]$$

2. Approach

This is taken from the solutions.

$$\dfrac{8}{3}cb\int_0^a\left[1-(\tfrac{x}{a})^2\right]^\frac{3}{2}dx \stackrel{x = az}{=} \dfrac{8}{3}cba\int_0^1\left[1-z^2\right]^\frac{3}{2}dz$$

where $\int_0^2\left[1-z^2\right]^\frac{3}{2}dz$ was given with $\dfrac{3}{16}\pi$.

Questions

  1. Why does the first approach come to another solution? What rule did I break?

  2. What happened with the substitution $x = az$ in the second approach? How does the a in the upper bound turn to 1 and why is $dz \neq \dfrac{1}{a}$.

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For APPROACH 1:

There were two errors in Approach 1.

First, $dx= a\,\cos z$ and not $\frac1{a\cos z}$.

Second, the upper limit transforms from $x=a$ to $z=\pi/2$.

Then, the integral of interest becomes

$$\frac83 abc\int_0^{\pi/2}\cos^4 z\,dz=\frac83 abc\times \frac{3\pi}{16}=\frac{\pi}{2}abc$$


For APPROACH 2:

Here, the substitution $x=az\implies dx=adz$ and when the upper limit on $x$ is $a$, the corresponding upper limit on $z$ is, therefore, $1$ since $a \times 1=a$.

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  • $\begingroup$ Could you give more explanation on how $x = a$ to $z = \pi/2$? If $x = a\sin(z)$ then we would need calculate the upper limit with $z = a \iff a\sin(a)$? Same argumentation for the limits in the second approach. $\endgroup$ – bodokaiser Aug 29 '15 at 7:01
  • $\begingroup$ Sure. So if $x_{upper}=a$ and $x_{upper}=a\sin z_{upper}$, then can you see why $z_{upper}=\pi/2$? Let's solve for $\sin z_{upper}$. It will be $1$. And the sine of $\pi/2$ is indeed $1$. Make sense? $\endgroup$ – Mark Viola Aug 29 '15 at 7:03
  • $\begingroup$ Ah okay I see. But when I normally use substitution e.g. $\int_0^1 (1-x)^\frac{2}{3}dx$ with $z = 1-x$ then this would yield $x_{upper} = 1-1 = 0$ (what I mean in this case we just "insert" the old limit into the substitution. Your approach tries to find a way how to express the old limit with the new substitution. Why is there this difference? $\endgroup$ – bodokaiser Aug 29 '15 at 7:09
  • $\begingroup$ Ah I now got the difference. You can do integration by substitution "from the left" or "from the right" this means you either completely substitute e.g. $u = 2x$ or you "wrap the substitution" e.g. $x = \sin u$. Depending on the exact substitution limit transformation must be done differently. I was now stuck the the first type of substitution. $\endgroup$ – bodokaiser Aug 29 '15 at 7:45

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