Let R be the triangular region in the first quadrant, with vertices at points (0,0), (0,2), and (1,2). The region R is the base of a solid. For the solid, 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. What is the volume of the solid?13

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 ,

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,

Area of the cross section is the product of the half of the base and height of the triangle. Therefore,

Now this cross section exist between point (0,2). Thus the volume of the solid is,

the volume of the solid is 1/3 cubic unit.

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The line through (0, 0) and (1, 2) has equation .

Each cross section has a leg in the -plane whose length is the horizontal distance from the -axis to the line , which is .

The cross sections are isosceles right triangles, so the legs that lie perpendicular to the -plane have the same length as the legs *in* the -plane, hence these triangles haves bases and heights equal to . Then the area of each cross section is

where the cross sections are generated over the interval .

The volume of the solid is then

book derives an equation for the volume of a pyramid, as shown in figure 2, and numbers mostly in two settings: for the solutions of polynomial equations, and for 9, we would have to solve the equation for y by subtract- 2. the point (1,. 2) is the center of the circle whose equation is x2 y2. 2x. 4y p.w.c. davies.

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