Finding Volume of Solid: Isosceles Right Triangle Cross-Sections

[Lo.Lee.Ta.]

Hi, I'm still practicing how to find volume.

1. My problem is this:

"Find the volume of the solid described below:

The base of the solid is the disk x^2 + y^2 ≤ 4. The cross-sections by planes perpendicular to the y-axis between y=-2 and y=2 are isosceles right triangles with one leg in the disk."
2. I thought that since an isosceles right triangle is really just half of a square, I could appraoch this problem like every cross-section was a square and divide the volume in half at the end...

Using this method I have:

A = s^2

A= (2√(4-y^2))^2

A= 4(4-y^2)

A= 16 - 4y^2

This is the area if every cross-section was a square.V= 1/2 * 2 ∫0 to 2 of 16 - 4y^2 (The 1/2 on the outside is there because we need to find half of the volume to end up with isosceles right triangle cross-sections. The 2 is there because I changed the limits from -2 to 2 instead to 0 to 2.)

V= ∫0 to 2 of 16 - 4y^2

= 16y - 4*(y^3/3) |0 to 2

= 16(2) - 4*[(2)^3)/3]

= 32 - 10.67

= 21.33

But this is the wrong answer... Why is it wrong, though?

I avoided having to think about the cross-sections as triangles, because the area formula for that is:
1/2(b*h)

I am confused as to what to plug in for the b and h...
I know for an isoscleses right triangle, h = 1/2b
Would b = sqrt(-y^2 + 4)...?
And so Area= 1/2[sqrt(-y^2 + 4) * (1/2(sqrt(-y^2 + 4)))]

...I really don't know here. That's why I avoided triangles!
Please help!
Thank you VERY much! :)

 

[HallsofIvy]

What makes you think the answer is wrong? Everything you have done is correct, although I would prefer you leave the answer as 32- 32/3= (96- 32)/3= 64/3, the exact answer, rather than 21.33, an approximate answer.

 

[haruspex]

A= (2√(4-y^2))^2

Not quite. Think that through again.

 

[Lo.Lee.Ta.]

@haruspex- I still don't see why that would be wrong...

Wouldn't each side be (2*(√4 - y^2))^2 ~ a (√(4 - y^2)) for the top part of the circle and a (√(4 - y^2)) for the bottom part of the circle?

 

[haruspex]

Maybe I have it wrong, but I thought that the x value, √(4 - y^2), would be half a diagonal of the complete square.

 

[Lo.Lee.Ta.]

Oh, okay. Thanks for responding! :)

 

[Dick]

Maybe I have it wrong, but I thought that the x value, √(4 - y^2), would be half a diagonal of the complete square.

Right. It said a leg of the isoceles triangle is in the disk. If you are dealing with an online grading system it might want the exact answer as a fraction instead of a decimal approximation as Halls pointed out.