Question

find the surface area of the object obtained by rotating y=4+3x^2, 1<=x<=2 about the y axis

Answer 1

$y=4+3x^2$ , $1\leq x\leq2$

Express x as function of y:

$x=\frac{1}{\sqrt3}\sqrt{y-4}$ , $7\leq y\leq 16$

The surface area can be calculated using the formula:

$S=2\pi\int_7^{16} x(y)\sqrt{1+(x'(y))^2}dy$

$x'=\frac{1}{2\sqrt3}\frac{1}{\sqrt{y-4}}$

$S=2\pi\int_7^{16} \frac{1}{\sqrt3}\sqrt{y-4}\sqrt{1+\frac{1}{12}\frac{1}{y-4}}dy=$$2\pi\int_7^{16} \frac{1}{\sqrt3}\sqrt{y-4}\sqrt{1+\frac{1}{12}\frac{1}{y-4}}dy=$$\frac{1}{3}\pi\int_7^{16} \sqrt{12y-47}dy=\frac{1}{3}\pi\cdot\frac{2}{3\cdot12}(12y-47)^{3/2}|_7^{16}=$$\frac{1}{3}\pi\cdot\frac{2}{3\cdot12}(12y-47)^{3/2}|_7^{16}=\frac{\pi}{54}(145^{3/2}-37^{3/2})$$\approx88.49$