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bleakarcher · Oct 5, 2011

http://www.angelfire.com/ab7/fourunit/sydneytech2004.pdf
q7) b) iii)

4025808 · Oct 5, 2011

The ellipse at point h (small h) is $(x^2/a^2\lambda^2) + (y^2/b^2\lambda^2) = 1$

did you get that?

the next part is using the fact that the area of an ellipse is $\pi(a)(b)$ , we subsitute the points a = $a\lambda$ and b = $b\lambda$
then multply we get area = $\pi a b \lambda^2$

then substitute lambda = 1 - (h^2/H^2)

then integrate from 0 to H with respect to dh (small h)

roryclifford · Oct 5, 2011

4025808 said:
The ellipse at point h (small h) is $(x^2/a^2\lambda^2) + (y^2/b^2\lambda^2) = 1$

did you get that?

the next part is using the fact that the area of an ellipse is $\pi(a)(b)$ , we subsitute the points a = $a\lambda$ and b = $b\lambda$
then multply we get area = $\pi a b \lambda^2$

then substitute lambda = 1 - (h^2/H^2)

then integrate from 0 to H with respect to dh (small h)

I just did this question, followed the same steps as you wrote here and got 8abpieH / 3, not / 15
Haha can you lead me through? I can't see any mistakes in my working

4025808 · Oct 5, 2011

roryclifford said:
I just did this question, followed the same steps as you wrote here and got 8abpieH / 3, not / 15
Haha can you lead me through? I can't see any mistakes in my working

sure thing

from my last step, I have it as

$V = \pi a b (1 - \frac{h^2}{H^2})^2$

$V = \pi a b (1 - 2\frac{h^2}{H^2} + \frac{h^4}{H^4})$

now you integrate from 0 to H

so $V = \pi a b \int_{0}^{H}(1 - 2\frac{h^2}{H^2} + \frac{h^4}{H^4}) dh$

and this results in

$V = \pi a b [h - \frac{2h^3}{3H^2} + \frac{h^5}{5H^4}]_{0}^{H}$

$V = \pi a b [H - \frac{2H^3}{3H^2} + \frac{H^5}{5H^4}]$

$V = \pi a b H [1 - \frac{2}{3} + \frac{1}{5}]$ , factorizing H outside

$V = \pi a b H [\frac{8}{15}]$

maybe you did something wrong with the fractions at the end?

bleakarcher · Oct 6, 2011

thank 4025808 lol, this was an extremely simple question but i didnt realise u had to divide by lambda for some reason lol

4025808 · Oct 6, 2011

bleakarcher said:
thank 4025808 lol, this was an extremely simple question but i didnt realise u had to divide by lambda for some reason lol

well the equation of the ellipse is always = 1, so you must divide by lambda to get 1 :L

bleakarcher · Oct 6, 2011

it was 12 in the night ok lol?

roryclifford · Oct 7, 2011

4025808 said:
sure thing

from my last step, I have it as

$V = \pi a b (1 - \frac{h^2}{H^2})^2$

$V = \pi a b (1 - 2\frac{h^2}{H^2} + \frac{h^4}{H^4})$

now you integrate from 0 to H

so $V = \pi a b \int_{0}^{H}(1 - 2\frac{h^2}{H^2} + \frac{h^4}{H^4}) dh$

and this results in

$V = \pi a b [h - \frac{2h^3}{3H^2} + \frac{h^5}{5H^4}]_{0}^{H}$

$V = \pi a b [H - \frac{2H^3}{3H^2} + \frac{H^5}{5H^4}]$

$V = \pi a b H [1 - \frac{2}{3} + \frac{1}{5}]$ , factorizing H outside

$V = \pi a b H [\frac{8}{15}]$

maybe you did something wrong with the fractions at the end?

Yeah I integrated the last term wrong haha, such a silly mistake.
Cheers for your solution

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