Intersection between the solid and a plane perpendicular

[Telemachus]

Homework Statement

A solid has a circular base of radius r=4. Find in each case the volume of the solid if all intersection between the solid and a plane perpendicular to a set diameter is:

a) A square
b) A triangle rectangle isosceles which one of its sides is on the circular base.

The Attempt at a Solution

The first I thought was in some kind of cylindroid. I know I must find an equation that I can integrate. So, the sides of the squares will be given by:

$$z=2\sqrt{r^2-x^2}$$

Is the integration given by: $$\displaystyle\int_{-4}^{4}[2\sqrt{r^2-x^2}]^2dx=4\displaystyle\int_{-4}^{4}(r^2-x^2)dx$$ for the square?

And how can I find the volume of the other solid b)?

 

[Dick]

That looks fine to me. Ok, so the area of a square with side z is z^2. And you integrated the area to get the volume. Fine. What's the area of a isosceles right triangle with side z? I'm not sure why this is confusing you. Are you worried about Odysseus' return?

 

[Telemachus]

Haha I still drunk from Blooms day :p

The problem I got is... the area of the triangle: $$\diplaystyle\frac{bh}{2}$$
How do I get h? Pythagoras?

 

[Dick]

Haha I still drunk from Blooms day :p

The problem I got is... the area of the triangle: $$diplayfrac\style{bh}{2}$$
How do I get h? Pythagoras?

In the triangle I'm thinking of both b and h are z. In an isosceles right triangle both sides are equal, aren't they?

 

[Telemachus]

If both of them are equal one each other, then they are not equal to the base. So I must find h...

 

[Dick]

If both of them are equal one each other, then they are not equal to the base. So I must find h...

There are two different ways to draw the picture. The question says 'one of its sides is on the base'. I took this to mean that it isn't the hypotenuse that has length z, but rather one of the legs. I'm probably wrong. If I am, yes, use pythagoras to find the missing length.

 

[Telemachus]

Sorry, I didn't give a righteous interpretation at what you said in the first place. You're right, it's easier to think of both, the base and the high as the information I already get to. So, it won't be hard to find the volume from there.

Thanks Dick.