Volume of a triangular based solid

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Homework Statement


the base of a solid is the triangular region bounded by the line x=0 (on the y-axis), the line y = 0 (the x-axis), and the line y = −2x + 2. Cross-sections perpendicular to the
y-axis are squares. Find the volume of S.

Homework Equations





The Attempt at a Solution


I was not sure where to start with this question. Is it something where is should be using disk shell method and rotating about an axis in order to find a volume? or is it done by finding the area of the triangular base (which I found to be 1) and multiplying by a length? I would really appreciate some help on how I could picture this question and go about it.
 
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spaniks said:

Homework Statement


the base of a solid is the triangular region bounded by the line x=0 (on the y-axis), the line y = 0 (the x-axis), and the line y = −2x + 2. Cross-sections perpendicular to the
y-axis are squares. Find the volume of S.

Homework Equations





The Attempt at a Solution


I was not sure where to start with this question. Is it something where is should be using disk shell method and rotating about an axis in order to find a volume? or is it done by finding the area of the triangular base (which I found to be 1) and multiplying by a length? I would really appreciate some help on how I could picture this question and go about it.
There's no rotation involved. Sketch the solid first to get an idea of what it looks like. Use the fact that cross sections perpendicular to the y-axis are squares to get the incremental volume, and then sum those volumes to get the total volume.
 
No, this is not rotated about an axis nor is it a volume. What you can do is argue that, slicing perpendicular to the y-axis, the volume of each such slice is its area times dy, then integrate with respect to dy.

I chose the y- axis because we are told that the cross sections perpendicular to the y-axis are squares. All you need to do is determine what the side length of each such square is. And a side of each square is from the y-axis to the line y= -2x+ 2. What is the length of that line segment in terms of y?
 
so if I solved y=-2x+2 for x which would be x=-1/2y+1 it would give me the length of one of the squares and then if I squared it and took the integral I could find the volume?
 
Yes. And you integrate from y = 0 to y = 2.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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