94.55π is the area of the required surface.
[tex]S=\int\limits^{x_1}_{x_2} {2\pi f(x)\sqrt{1+f'(x)^2} \, dx[/tex]
is the exact area of a surface obtained by rotating the curve y=f(x), [tex]x_1\leq x\leq x_2[/tex] about the x-axis
Given,
[tex]x=\frac{1}{3}(y^2+2)^{\frac{3}{2}}[/tex]
[tex]\implies (3x)^\frac{2}{3}=y^2+2\\\implies y^2=(3x)^\frac{2}{3}-2[/tex]
[tex]\implies y=\sqrt{(3x)^\frac{2}{3}-2}[/tex]
Let
[tex]y=\sqrt{(3x)^\frac{2}{3}-2}=f(x)[/tex]
Now, the range of y is given 3≤y≤4
hence, the range of x is [tex](\frac{1}{3}(3^2+2)^{\frac{3}{2}}[/tex],[tex]\frac{1}{3}(4^2+2)^{\frac{3}{2}})[/tex]=(12.16,25.46).
So, [tex]S=\int\limits^{25.46}_{12.16} {2\pi f(x)\sqrt{1+f'(x)^2} dx\\\\=[\frac{\pi}{2}(3x)^{\frac{4}{3}}-\pi (3x)^{\frac{2}{3}}]^{25.46}_{12.16}\\=101.63\pi - 6.99 \pi \\= 94.55 \pi[/tex]
is the area of the required surface.
Hence, 94.55π is the area of the required surface.
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