Finding Volume of Solids Bounded by Curves

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PWNC89

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I've been having trouble with two Calculus problems, and I'm really confused as to where I'm supposed to start.

Problem 1 The base of the solid S is bounded by the curves y = sqrt x and y = 2x. Its cross-sections perpindicular to the x-axis are equilateral triangles. Find the volume of S.

All I have for this one is that the area of the triangle is ((sqrt 3) / 4 ) s^2

Problem 2 The region R is bounded by the curves y =x^2 + 4, y = x^3 and x = 0. Find the volume of the solid generated by rotating R about the x-axis.

Any help is greatly appreciated.
 
\(\displaystyle 1) \ V \ = \ \frac{\sqrt3}{4}\int_{0}^{1/4}(x^{1/2}-2x)^{2}dx\)


\(\displaystyle 2) \ V \ = \ \pi\int_{0}^{2}[(x^{2}+4)^{2}-(x^{3})^{2}]dx\)
 
Thanks! After using what you gave me I came up with some ridiculous answers. Would you mind looking over my work for any mistakes?
 
You have to know basic algebra before you advance to the next plateau.

What is [x^(1/2)-2x]^2 expanded?
 
(x^(1/2) - 2x)(x^(1/2) - 2x)
When you multipy variables with exponents you add the exponents, no?
Therefore by FOIL method:
x - 2x^-(1/2) - 2x^-(1/2) + 4x^2
 
\(\displaystyle Look \ pal \ (x^{1/2}-2x)^{2} \ = \ x-4x^{3/2}+4x^{2}.\)

\(\displaystyle x \ = \ x^{1} \ = \ x^{2/2}, \ the \ one \ is \ understood \ and \ usually \ not \ emphasized.\)
 
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