The triangle with vertices (0,0) (0,3) and (2,0) is rotated about the yaxis. find the volume.?

Question

it says use the disk method THEN the shells..

thanks in advance for any help!

Answer 1

The formula for the volume of the cone is helpful for checking whether one is correct.
The Disc Method:
We are summing all the little discs which can approximate this figure, allowing the thickness of each one to become infinitessimal. The volume of a disc is its base area multiplied by thickness. The area is determined by the radius, which is 'x' in this case. We cannot use 'x', however, because the integration is happening along the y-axis, so:
The hypoteneus of your triangle is characterised by the line:
y = -3/2x + 3, solve for x to get:
x = -2/3y + 2.
Now,
A = πr², i.e.,
A = π( -2/3y + 2 )²

You can multiply by the infinitessimal, dy, and integrate from 0 to 3:
integral(π( -2/3y + 2 )²)dy, [0,3];
π*integral(( -2/3y + 2 )²)dy, [0,3];
π*integral( -4/9y² - 8/3y + 4 )dy, [0,3];

π*[-4/27y³ - 8/6y² + 4y][0,3]
which is,
π*(0 -(-4/27*27 -8/6*9 + 4*3)),
π*(0 -(-4 - 12 + 12)),
π*4.

Best to use a calculator here to check...

I'm a bit rusty with the shells method, at least to explain it vigorously. You can check out
http://en.wikipedia.org/wiki/Shell_integration
to get started.

Answer 2

This is just a right angle cone with radius = 2 and Height =3. Volume = PI r^2 * h/3 = 3.14*2*2*3/3 = 12.56 Not sure about how disks and shells apply here.

Answer 3

disk method: π∫ f(y)^2dy
you're basically taking the area of circles of radius x=f(y) and adding them all up as you go up the y-axis (the limits of y will be your limits). If you were rotating aout the x-axis, the integral would be in terms of y=f(x) and the limits of x would denote your limits.

shell method: 2π∫ xf(x)dx
this time, you're making cylinders and adding up their lateral surface areas. Lateral surface area for cylinder = 2πrh, or circumference x height, so this time, you need 2 functions: one for radius and one for height. You'll use the variable whose axis is perpendicular to the axis of rotation for the integration variable and the limits of that variable for the limits of the integration.

simplest method (triple integral): ∫∫∫dydrdθ
if you want you can just integrate the cone directly with a triple (volume) integral. All you have to do is pick the right limits and integrate the function "1". In this case, you could pick 0

Answer 4

using disks, it's
integral from 0 to 3 of pi f(y)^2 dy
where f(y) is given by the line (d) from (0,3) to (2,0).
This line has equation y - 0 = (-3/2)( x- 3), so
x = (-2/3) y +2 .
so integral from 0 to 3 of (pi(-2/3)y +2)^2 dy)

using shells:
integral fom 0 to 2 of 2 pi g(x) dx
where g(x) is given from the above equation of (d)
g(x) = (-2/3) (x - 3)