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 \(0\le x\le3\).
\(A_{\Delta}=\frac{a^2\sqrt{3}}{4}\)
\(\large\frac{k^2\sqrt{3}}{4}\\ \Rightarrow \ k_0^1=2\cdot f(x)\\ \Rightarrow \ k_1^3=2\cdot g(x)\)
\({\color{black}x=y^2\\ \Rightarrow} \ f(x)=\sqrt{x}\\ {\color{black}2y^2=3-x\\ \Rightarrow} \ g(x)=\sqrt{\frac{3-x}{2}}\)
\(V=\int_{0}^{1} \! (2\cdot f(x))^2\cdot \frac{\sqrt{3}}{4} \, dx +\int_{1}^{3} \! (2\cdot g(x))^2\cdot \frac{\sqrt{3}}{4} \, dx \)
\(V=\int_{0}^{1} \! (2\cdot \sqrt{x})^2\cdot \frac{\sqrt{3}}{4} \, dx +\int_{1}^{3} \! (2\cdot \sqrt{\frac{3-x}{2}})^2\cdot \frac{\sqrt{3}}{4} \, dx \)
\(V=\sqrt{3}(\int_{0}^{1} \! x \, dx +\frac{1}{2}\int_{1}^{3} \! (3-x) \, dx )\)
\(V=\sqrt{3}\cdot (|\frac{x^2}{2}|_0^1+|3x-\frac{x^2}{2}|_1^3)\\ V=\sqrt{3}\cdot (\frac{1}{2}+9-\frac{9}{2}-(3-\frac{1}{2})\)
\(V=\frac{5\cdot \sqrt{3}}{2}=4.33\)
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