Author Topic: Newington 2003- igloo question volumes (Read 1446 times) Tweet Share

massive · « **on:** August 16, 2016, 10:28:32 pm »

Hey does anyone know how to do part iv of this igloo question? I don't understand how you find the area of the cross section. :/

RuiAce · « **Reply #1 on:** August 17, 2016, 12:13:29 am »

Quote from: massive on August 16, 2016, 10:28:32 pm

Hey does anyone know how to do part iv of this igloo question? I don't understand how you find the area of the cross section. :/

$\text{Consider the }x\text{-}y\text{ plane. If we make }y\text{ the subject of the equation of the ellipse:}\\ y=\pm \sqrt{\frac{4-x^2}{4}}=\pm \frac{\sqrt{4-x^2}}{2}$
$\text{Consider the }y\text{-}z\text{ plane.}\\ \text{You need to convince yourself that the parabolic cross section has EQUATION }z=h-y^2 \\ \text{Because we know what }h\text{ is equal to we have }\\ z=\frac{1}{2}(4-x^2)-y^2$
$\text{The area bound between the curve (parabola) and the axis can be calculated as the following integral:}\\ A=\int_{-y}^{y} \left(\frac{1}{2}(4-x^2)-y^2\right)dy \\ \text{But from what I said at the start, we know that the boundaries are }\pm y = \pm \frac{\sqrt{4-x^2}}{2}\\ \text{!! Do you see why?}$
$\begin{align*}A&=\int_{-\frac{\sqrt{4-x^2}}{2}}^{\frac{\sqrt{4-x^2}}{2}}\left(\frac{1}{2}(4-x^2)-y^2\right)dy \\ &= 2\int_{0}^{\frac{\sqrt{4-x^2}}{2}}\left(\frac{1}{2}(4-x^2)-y^2\right)dy \tag{even function integral}\\ &= 2\left[\frac{1}{2}(4-x^2)y-\frac{y^3}{3}\right]_{0}^{\frac{\sqrt{4-x^2}}{2}}\end{align*}$

Yeah I made a mistake somewhere but I'm too tired to figure it out. The coefficients don't add to 1/3

RuiAce · « **Reply #2 on:** August 17, 2016, 08:31:30 am »

massive · « **Reply #3 on:** August 17, 2016, 10:51:08 am »

Damn thank you so much man!!

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