# Finding Volume of Solid: Isosceles Right Triangle Cross-Sections

• Lo.Lee.Ta.
In summary, my problem was that I thought I could divide the volume of the solid by dividing the area of the cross-sections by squares instead of triangles. I got the wrong answer because I didn't take into account that an isosceles right triangle is really just half of a square and the area of a square is 1/2(b*h).
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!
Thank you VERY much! :)

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.

Lo.Lee.Ta. said:
A= (2√(4-y^2))^2
Not quite. Think that through again.

@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?

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

Oh, okay. Thanks for responding! :)

haruspex said:
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.

## What is an isosceles right triangle?

An isosceles right triangle is a triangle with two sides of equal length and one right angle.

## How do you find the volume of a solid with isosceles right triangle cross-sections?

To find the volume of a solid with isosceles right triangle cross-sections, you can use the formula V = (1/3) * b * h * l, where b is the base length of the triangle, h is the height of the triangle, and l is the length of the solid.

## What is the relationship between the cross-sectional area and the volume of a solid with isosceles right triangle cross-sections?

The cross-sectional area of an isosceles right triangle is directly proportional to the volume of the solid. This means that as the cross-sectional area increases, the volume also increases.

## Can the volume of a solid with isosceles right triangle cross-sections be negative?

No, the volume of a solid cannot be negative. It represents the amount of space occupied by the solid and must always be a positive value.

## What are some real-life applications of finding the volume of solids with isosceles right triangle cross-sections?

One real-life application is in construction, where finding the volume of a triangular prism can help determine the amount of material needed for a project. Another application is in 3D printing, where the volume of a solid object with isosceles right triangle cross-sections can be used to calculate the amount of material required for printing.

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