A region bounded

Question

Region R is bounded by  and the x-axis over the interval [–1, 1]. A solid has base R, and cross sections perpendicular to the x-axis are right isosceles triangles with one leg located in the base. The volume of this solid is

.

.

.

.

Answer 1

First calculate the area A(x) of the solid's cross-section at an arbitrary position x.

It is a right isosceles triangle with base y = cos(πx/2)/2.

Therefore the area of that triangle is

A(x) = y²/2 = cos²(πx/2)/8.

Later, for the purposes of integration it will turn out convenient to rewrite

A(x) using the identity

cos(2a) = 2cos²(a) - 1

Thus

cos²(a) = (cos(2a) + 1)/2

Therefore

A(x) = (cos(πx) + 1)/16

The volume of the solid is then the sum of all those triangles.

That is the definite integral from -1 to 1 ∫A(x)dx

Let F(x) = ∫A(x)dx be the indefinite integral.

Once we figure out F(x), the answer to the question will be F(1) - F(-1).

F(x) = ∫A(x)dx =

(1/16)∫(cos(πx) + 1) dx =

(1/16)(sin(πx)/π + x)

F(1) = (1/16)(0 + 1) = 1/16

F(-1) = (1/16)(0 - 1) = -1/16

The volume of the solid is 1/8.