volumes help! (1 Viewer)

[jkwii]

got a Q from patel.

the base of a solid is 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.

 

[Affinity]

at x the cross section has area

(2x/3)sqrt(8x - x^2)

intgrate from 0 to 8

x=v+4 might be a good substitution

 

[YannY]

got a Q from patel.

the base of a solid is 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.

The base: 2y

Lets find the Height: one third of the distance of the plane of the section from the origin. i.e x/3

The circle: x^2 + y^2 = 8x
(x-4)^2+y^2=16
Thus, the bounds: 0 - 8

V= Integral (Area)
= I (x/3.2y)dx
= 2/3I (xsqrt(16-(x-4)^2)dx
= 64pi/3

[Sorry cnt be fked integrating this, i did it on a graph and it has the same value as 64pi/3, hahaha]

 

[YannY]

Ah what the heck,

xsqrt(16-(x-4)^2)

let x-4=4sin@
dx=4cos@d@
when x=8 @=pi/2
x=0 @=-pi/2

therefore I=(4sin@+4)sqrt(16-16sin@)4cos@d@ with bounds pi/2, -pi/2
=64sin@cos^2@+64cos^2@
=64sin@cos^2@+32+32cos2@
=-64/3.cos^3@+32@+16sin2@
=32pi

However, the original integral had 2/3 infront hence
2/3 x 32pi = 64pi/3units^3

 

[namburger]

yanny, Volumes already..

 

[YannY]

Dw namburger, im doing volumes means im not doing integration + graphing atm thus not striking the iron while its hot.

So dont feel so bad, you'll cane me in the next test this way .

 

[jkwii]

got another one argh!!!

a certain soldi has a semi-circular base of radius 4 units. the cross sections at right angles to the x axis are triangles with one leg in the base. if the height of these triangles are bounded by the arc of the parabola y=16-x^2 show that the volume of the solid is given by:

bounds are 0-8

V=I (16-x^2)^3/2 dx and hence find V