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Timothy.Siu · Feb 5, 2009

The base of a solid is the circle x^2+y^2=8x and every plane section perpendicular to the x-axis is a rectangle whose height is one third of the distance of the plane of the section from the origin. Show that the volume of the solid is 64pi/3.

independantz · Feb 5, 2009

$x^2+y^2=8x\\x^2-8x+16+y^2=16\\(x-4)^2+y^2=16\\ \text{ draw the thing if you like}\\ \text{a Typical slice: } \Delta v= 2xy/3\\\therefore v=\lim_{\Delta x->0}\sum_{x=0}^{8}2xy/3.\Delta x\\=\lim_{\Delta x->0}\sum_{x=0}^{8}=2x/3\sqrt{16-(x-4)^2}.\Delta x \text{,By rearranging the equation of the circle}\\=2/3\int_{0}^{8}x.\sqrt{16-(x-4)^2}dx\\\text{Now let: } x-4=4sin\Theta\\\therefore dx=4cos\theta.d\theta\\\text{When: }x=8,\theta=-pi/2\\\text{When: }x=0, \theta=-pi/2\\\therefore v=2/3\int_{-pi/2}^{pi/2}(4sin\theta+4)\sqrt{16-16sin^2\theta}.4cos\theta.d\theta\\=18/3\int_{-pi/2}^{pi/2}16sin\theta cos^2\theta.d\theta+8/3\int_{-pi/2}^{pi/2}16cos^2\theta.d\theta\\=-128/3[cos^3\theta/3]\binom{pi/2}{-pi/2}+128/3\int_{-pi/2}^{pi/2}1/2[1+cos2\theta].d\theta\\=-128/3[0]+128/3[\theta/2]\binom{pi/2}{-pi/2}+128/12[sin2\theta]\binom{pi/2}{-pi/2}=64pi/3 \text{ units}^2$

I skipped a few lines as it took to long to type, however if you need any further assistance just post

Timothy.Siu · Feb 5, 2009

yeah thanks a lot
i got most of it,
i see where i went wrong now.

independantz · Feb 6, 2009

superelizabeth said:
so this is what's waiting for me in Yr 12 ..... *faints* T_T

Lol, Don't worry it's not as hard as it looks

tommykins · Feb 6, 2009

superelizabeth said:
so this is what's waiting for me in Yr 12 ..... *faints* T_T

don't worry, keep up with your algebra and you'll be fine.

untouchablecuz · Feb 7, 2009

independantz said:
$x^2+y^2=8x\\x^2-8x+16+y^2=16\\(x-4)^2+y^2=16\\ \text{ draw the thing if you like}\\ \text{a Typical slice: } \Delta v= 2xy/3\\\therefore v=\lim_{\Delta x->0}\sum_{x=0}^{8}2xy/3.\Delta x\\=\lim_{\Delta x->0}\sum_{x=0}^{8}=2x/3\sqrt{16-(x-4)^2}.\Delta x \text{,By rearranging the equation of the circle}\\=2/3\int_{0}^{8}x.\sqrt{16-(x-4)^2}dx\\\text{Now let: } x-4=4sin\Theta\\\therefore dx=4cos\theta.d\theta\\\text{When: }x=8,\theta=-pi/2\\\text{When: }x=0, \theta=-pi/2\\\therefore v=2/3\int_{-pi/2}^{pi/2}(4sin\theta+4)\sqrt{16-16sin^2\theta}.4cos\theta.d\theta\\=18/3\int_{-pi/2}^{pi/2}16sin\theta cos^2\theta.d\theta+8/3\int_{-pi/2}^{pi/2}16cos^2\theta.d\theta\\=-128/3[cos^3\theta/3]\binom{pi/2}{-pi/2}+128/3\int_{-pi/2}^{pi/2}1/2[1+cos2\theta].d\theta\\=-128/3[0]+128/3[\theta/2]\binom{pi/2}{-pi/2}+128/12[sin2\theta]\binom{pi/2}{-pi/2}=64pi/3 \text{ units}^2$

I skipped a few lines as it took to long to type, however if you need any further assistance just post

OR you can use the substitution u = x-4

then evaluate the integral from the properties of integrals

the integral turns into

(2/3) int [4 -> -4] (u +4) sqrt(16-u^2)du =

2/3 int [4 -> -4] (u sqrt(16-u^2)du) + (8/3) int [4 -> -4] sqrt(16-u^2)du =

0 + 64pi/3

since u sqrt(16-u^2) is odd and the other integral is just half of a semicircle with radius 4

(sorry, i dont know how to use latex)

volumes q (1 Viewer)

Timothy.Siu

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independantz

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Timothy.Siu

Prophet 9

independantz

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tommykins

i am number -e^i*pi

untouchablecuz

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