The line through (0, 0) and (1, 2) has equation [tex]y=2x[/tex].
Each cross section has a leg in the [tex]xy[/tex]-plane whose length is the horizontal distance from the [tex]y[/tex]-axis to the line [tex]y=2x[/tex], which is [tex]x=\frac y2[/tex].
The cross sections are isosceles right triangles, so the legs that lie perpendicular to the [tex]xy[/tex]-plane have the same length as the legs *in* the [tex]xy[/tex]-plane, hence these triangles haves bases and heights equal to [tex]\frac y2[/tex]. Then the area of each cross section is
[tex]\dfrac12\left(\dfrac y2\right)^2=\dfrac{y^2}8[/tex]
where the cross sections are generated over the interval [tex]0\le y\le2[/tex].
The volume of the solid is then
[tex]\displaystyle\int_0^2\frac{y^2}8\,\mathrm dy=\frac1{24}y^3\bigg|_0^2=\frac{2^3}{24}=\frac13[/tex]
The volume of the solid is 1/3 cubic unit.
Given-
Triangular region is R.
Vertices of this region is at points (0,0) (0,2) and (1,2).
The equation for the line by which the point (0,0) and (1,2) passes can be given as ,
[tex]y=2x[/tex]
[tex]x=\dfrac{y}{2}[/tex]
We have given that each cross section perpendicular to the y-axis is an isosceles right triangle with the right angle on the y-axis and one leg in the xy-plane. Hence, the height and base of the triangle are equal which is,
[tex]\dfrac{y}{2}[/tex]
Area of the cross section is the product of the half of the base and height of the triangle. Therefore,
[tex]A=\dfrac{1}{2} \times \dfrac{y}{2} \times \dfrac{y}{2}[/tex]
[tex]A=\dfrac{y^2}{8}[/tex]
Now this cross section exist between point (0,2). Thus the volume of the solid is,
[tex]V=\int\limits^2_0 {\dfrac{y^2}{8}} \, dy[/tex]
[tex]V=\dfrac{2^3}{24}[/tex]
[tex]V=\dfrac{1}{3}[/tex]
the volume of the solid is 1/3 cubic unit.
For more about the volume, follow the link below
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