Answer to Question #249725 in Calculus for SHAIKAT

Question

Find the area of the surface that is generated by revolving the portion of the curve y = x

2

between x = 2 and x = 3 about the x-axis.

Answer 1

$$S=\displaystyle\int_{2}^{3}2\pi y\sqrt{1+(y')^2}dx$$

$$=\displaystyle\int_{2}^{3}2\pi x^2\sqrt{1+(2x)^2}dx$$

$$\int x^2\sqrt{1+(2x)^2}dx=$$

$$x=\dfrac{1}{2}\tan \theta, -\dfrac{\pi}{2}<\theta <\dfrac{\pi}{2}$$

$$dx=\dfrac{1}{2\cos^2\theta}d\theta$$

$$\sqrt{1+(2x)^2}=\dfrac{1}{\cos\theta}$$

$$\int x^2\sqrt{1+(2x)^2}dx=2\int (x^2)\sqrt{1+(2x)^2}dx$$

$$=\int \dfrac{1}{4}\tan^2 \theta(\dfrac{1}{\cos\theta})(\dfrac{1}{2\cos^2\theta})d\theta$$

$$=\dfrac{1}{8}\int (\dfrac{1}{\cos^3\theta}-\dfrac{1}{\cos^5\theta})d\theta$$

$$=\dfrac{1}{8}(\ln(|\tan\theta+\sec \theta|)+\tan\theta\sec\theta(1-2\sec^2\theta))+C$$

$$=\dfrac{\sqrt{1+4x^2}(16x^3+2x)-\ln(|\sqrt{1+4x^2}+2x|)}{64}+C$$

$$S=\dfrac{\pi}{32}\bigg(438\sqrt{37}-\ln(\sqrt{37}+6)$$

$$-132\sqrt{17}+\ln(\sqrt{17}+4)\bigg)\ square\ units$$