Calc Help!

I'll TRY.....First here is a graph of the area of the base:

  area of equilateral triangle =  sqrt(3)/4 s^2

s in this case = 2x from 0 to 3

 from 0 to 1   this is    2y^2            from  1 to 3 this is     6-4y^2      now square s

                                  4 y⁴                                                     16y^4 -48y^2 + 36

sqrt( 3)/4  ( 4y^4 )                             +          sqrt (3)/4 ( 16 y^4 - 48y^2 + 36)   

      this part is y 0 to 1                                          this part is y is 1 to 0

Hmmmm...... I am a little stuck...a bit rusty......If I figure this out (this may not be the EXACT way to do it) ...I'll edit this post

    maybe asinus will answer correctly....

The question is as follows:

The base of a solid is the region between the parabolas x=y^2 and 2y^2 = 3-x . Find the volume of the solid if the cross-sections perpendicular to the x-axis are equilateral triangles.

Hello Guest!

The parabolas intersect in P(1,1) and P(1,-1). They share the domain of the integral.

The entire domain is $0\le x\le3$.

$A_{\Delta}=\frac{a^2\sqrt{3}}{4}$       

$\large\frac{k^2\sqrt{3}}{4}\\ \Rightarrow \ k_0^1=2\cdot f(x)\\ \Rightarrow \ k_1^3=2\cdot g(x)$

${\color{black}x=y^2\\ \Rightarrow} \ f(x)=\sqrt{x}\\ {\color{black}2y^2=3-x\\ \Rightarrow} \ g(x)=\sqrt{\frac{3-x}{2}}$

$V=\int_{0}^{1} \! (2\cdot f(x))^2\cdot \frac{\sqrt{3}}{4} \, dx +\int_{1}^{3} \! (2\cdot g(x))^2\cdot \frac{\sqrt{3}}{4} \, dx$

$V=\int_{0}^{1} \! (2\cdot \sqrt{x})^2\cdot \frac{\sqrt{3}}{4} \, dx +\int_{1}^{3} \! (2\cdot \sqrt{\frac{3-x}{2}})^2\cdot \frac{\sqrt{3}}{4} \, dx$

$V=\sqrt{3}(\int_{0}^{1} \! x \, dx +\frac{1}{2}\int_{1}^{3} \! (3-x) \, dx )$

$V=\sqrt{3}\cdot (|\frac{x^2}{2}|_0^1+|3x-\frac{x^2}{2}|_1^3)\\ V=\sqrt{3}\cdot (\frac{1}{2}+9-\frac{9}{2}-(3-\frac{1}{2})$

$V=\frac{5\cdot \sqrt{3}}{2}=4.33$

  !