+0

# Calc Help!

0
113
3

Heyo, usually i dont like using Help Forums since I like doing things myself but im really stumped on how to solve this.

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.

A previous question uses a cross-sectional integral which is (k^2 * sqrt(3))/4 and you find k by subtracting the upper and lower bounds respectively, then you take  (k^2 * sqrt(3))/4 and find the integral of it for the bound of the shape. Using this method, I still cannot find the correct answer.

Feb 15, 2021

#1
+31826
+1

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

Feb 15, 2021
edited by ElectricPavlov  Feb 15, 2021
#3
+31826
0

s in this case = 2x from 0 to 3

should be      s in this case = 2 y from 0 to 3

ElectricPavlov  Feb 16, 2021
#2
+11438
+1

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 $$0\le x\le3$$.

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

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

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

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

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

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

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

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

!

Feb 15, 2021
edited by asinus  Feb 15, 2021
edited by asinus  Feb 15, 2021
edited by asinus  Feb 16, 2021
edited by asinus  Feb 16, 2021
edited by asinus  Feb 16, 2021