Finding a Volume using integration

[vipertongn]

Homework Statement

The base of a solid is a circle of radius 3. Find the volume of the solid if
parallel cross-sections perpendicular to the base are isosceles right triangles with
hypotenuse lying along the base.

Homework Equations

pi * Integral Rout^2- Rin^2

The Attempt at a Solution

I've tried to do the equation above but i believe I'm just having a problem visualizing this problem. Can someone run a step by step process in solving this equation.

I already know the answer is 36

My guess is like I find the area of triange and since the radius is 3 then that should be the hypotenuse...which then means the sides are 3/sqrt(2) the area becomes 3...and then the volume would be 2 integral of 3 from 0-6 which becomes 36

 

[rwisz]

Correction: I was not doing calculus, I was finding the area of a single rectangle and in no way, shape, or form finding volume by using known cross-sections.

Mark44's hint is what I would say now that I realize what I was trying to do!

 

[Mark44]

Both of you are way off on this one. The integral formula for "washers" doesn't apply here--this is not a solid of revolution. The radius of the circle is 3, but most of the triangles do not have a base length of 3.

It's helpful to draw a couple of pictures here--a 3D picture of the solid, and another of the base. Slice the solid up into triangular sections. Each triangular region extends from a point on the circle to the point across the x-axis from it, meaning that it extends from y = -sqrt(9 - x^2) to y = +sqrt(9 - x^2). The altitude of each of these triangular regions is half the base. The thickness of each region is $\Delta x$.

Is that enough of a hint? And yes, the volume comes out 36.

 

[vipertongn]

so it would be like a cylinderwith triangles in it? I can't visualize it. or maybe a cone...

 

[vipertongn]

So... the length of the hypotenuse is actually 6 therefore the sides should be 6/sqrt(2) and the height must be 36-18=18 so then tis sqrt(18)

i don't even know if I'm going at this correctly...

 

[Mark44]

The hypotenuse is 6 only in the center of the figure.

In slices away from the center, the hypotenuse gets smaller and smaller.

 

[vipertongn]

so i should use y^2=r^2-9? as the hypotenuse?

 

[Mark44]

so i should use y^2=r^2-9? as the hypotenuse?

Why would you use that?

 

[Mark44]

See post 3.

 

[vipertongn]

man I'm totally confused, i read the third post but I'm still not sure what's going on for this problem...

because the hypotenuse is lying on the base and the circle is the base I'm guessing that y^2=9-x^2 using the x^2+y^2=r^2

 

[vipertongn]

Both of you are way off on this one. The integral formula for "washers" doesn't apply here--this is not a solid of revolution. The radius of the circle is 3, but most of the triangles do not have a base length of 3.

It's helpful to draw a couple of pictures here--a 3D picture of the solid, and another of the base. Slice the solid up into triangular sections. Each triangular region extends from a point on the circle to the point across the x-axis from it, meaning that it extends from y = -sqrt(9 - x^2) to y = +sqrt(9 - x^2). The altitude of each of these triangular regions is half the base. The thickness of each region is $\Delta x$.

Is that enough of a hint? And yes, the volume comes out 36.

OH mark why is the altitude half the base?

 

[lanedance]

i think we need to step back to the question, though its not written fantatsically

Homework Statement

The base of a solid is a circle of radius 3. Find the volume of the solid if
parallel cross-sections perpendicular to the base are isosceles right triangles with
hypotenuse lying along the base.

this is not a cone, (soory for mis-steer) it would look like 2 curved surfaces meeting in a circular ridge above, and with a circular base. Any vertical slice perpidicular to the ridge is a right isoceles triangle.

This mean its top angle is pi/2 radians. So the base angles will both be pi/4

draw one!

Consider the triangles made by dividng this main one in half. They will both be the same with pi/4 at the main triangle peak and base boundary, with a pi/2 angle at the centre of the base of the main triangle. This shows the base of the colume is twice its height...

say we define (x,y,z) with (0,0,0) at the base of the volume, z vertical and x along the ridge direction..

how best to set up the integral? what is the easiest dV to integrate? well we know every vertical slice perpidicular to the x direction is a triangle, what are the height and width of the triangle in terms of x...?

then can you define an infintesimal volume based on this triangle and integrate over x, same as you would for a disk, washre or shell integration, except this one is in terms of triangles...

 

[Mark44]

man I'm totally confused, i read the third post but I'm still not sure what's going on for this problem...

because the hypotenuse is lying on the base and the circle is the base I'm guessing that y^2=9-x^2 using the x^2+y^2=r^2

That is correct, and is different from what you wrote a couple of posts ago.

so i should use y^2=r^2-9? as the hypotenuse?

You still need to solve for y, though.