1. ## Calculus Trouble

I am currently having trouble with these calculus practice problems. I would appreciate any help in solving these. Thank you.

1. Find the volume of a solid whose base is enclosed by the circle x^2 + y^2=1, and whose cross sections taken perpendicular to the base are equilateral triangles with bases in the xy-plane.

2. Let R be the region enclosed by the graphs of y=ln(x^2 + 1) and y=cos(x). Setup, using integrals, but do not evaluate the following:
a. The area of R.
b. The base of a solid is region R. Each cross section of the solid perpendicular to the x-axis is an isosceles right triangle. Express the volume of the solid.

2. Originally Posted by babolat29
I am currently having trouble with these calculus practice problems. I would appreciate any help in solving these. Thank you.

1. Find the volume of a solid whose base is enclosed by the circle x^2 + y^2=1, and whose cross sections taken perpendicular to the base are equilateral triangles with bases in the xy-plane.
The volume V of a solid between a and b is the integral of its cross-sectional area A(x) from a to b, or $\displaystyle \int_{a}^{b} A(x) dx$
The area A of an equilateral triangle is $\displaystyle \frac{\sqrt{3}s^2}{4}$ where $\displaystyle s$ is the length of a side. Since the trianglular cross-sections have their base in a circle with radius 1, the side of the triangular cross-section at any x is twice the distance from the x-axis to a semi-circle of radius 1, or $\displaystyle s(x) = 2\sqrt{1 - x^2}$. Therefore $\displaystyle A(x) = \frac{\sqrt{3}}{4}(2\sqrt{1 - x^2})^2 = \frac{\sqrt{3}}{4}4(1-x^2) = \sqrt{3}(1-x^2).$
Since your solid is enclosed in a circle with radius 1, you're going to integrate the area from -1 to 1, or $\displaystyle \int_{-1}^{1} A(x) dx = \int_{-1}^{1}\sqrt{3}(1-x^2)dx$ which you should know how to do.

3. For #2 See attachment