Posted by **Drake** on Tuesday, February 2, 2010 at 4:01am.

R is the region in the plane bounded below by the curve y=x^2 and above by the line y=1.

(a) Set up and evaluate an integral that gives the area of R.

(b) A solid has base R and the cross-sections of the solid perpendicular to the y-axis are squares. Find the volume of the solid.

(c) A solid has base R and the cross-sections of the solid perpendicular to the y-axis are equilateral triangles. Find the volume of the solid.

- Calculus -
**drwls**, Tuesday, February 2, 2010 at 4:19am
(a) The y = x^2 curve crosses y = 1 at x = 1. For the area in question, compute the integral of x^2 from x = 0 to 1, and subtract it from 1.

(b)Integrate along the y axis, with differential slab volume x^2 dy = y dy, from y=0 to y=1

(c) Proceed similary to (b), integrating along y from 0 to 1, but with equilateral triangle slabs. Slab volume must be expressed in terms of x

- Calculus -
**Drake**, Tuesday, February 2, 2010 at 11:17pm
Why would it only be the integral of x^2 and not the integral of 1-x^2 from x=-1 to x=1? So wouldn't that equal an area of 4/3? And secondly, how would you sketch these solids? when you say integrate along the y-axis do you mean to say that the curves have to be shifted as well? And why are all the integrals from x=0 to x=1?

- Calculus -
**caleb**, Tuesday, April 19, 2011 at 10:26pm
Find the volume, V , of the solid obtained

by rotating the region bounded by the graphs

of

x = y2, x = squareroot(y)

about the line x = −1.

## Answer This Question

## Related Questions

- Calculus - R is the region in the plane bounded below by the curve y=x^2 and ...
- Calculus - R is the region in the plane bounded below by the curve y=x^2 and ...
- calculus - Set up, but do not evaluate, the integral which gives the volume when...
- math - The region R, is bounded by the graphs of x = 5/3 y and the curve C given...
- math - The region R, is bounded by the graphs of x = 5/3 y and the curve C given...
- MATH-HELP! - The region R, is bounded by the graphs of x = 5/3 y and the curve C...
- why won't anybody help me - The region R, is bounded by the graphs of x = 5/3 y ...
- calculus - 1. Find the area of the region bounded by f(x)=x^2 +6x+9 and g(x)=5(x...
- Calculus AB - Let R be the region bounded by the graphs of y=sin(pi x) and y=(x^...
- Calculus check - The functions f and g are given by f(x)=sqrt(x^3) and g(x)=16-...

More Related Questions