Find the area of the surfaces of revolution for $y = 1 − x ^2 / 4$

Question

Find the area of surface of revolution obtained by revolving about the y-axis the curve

$$y = 1 − x ^2 / 4$$ from

$x = 0\,$ to $\,x = 2 √ 3$.

I really need help here. see my working on the attached image file below.

Answer 1

To find the surface of the of revolution obtained by revolving about the $y$-axis the parabola $y=1-\dfrac{x^2}{4}$in the interval $[0;\;2\sqrt 3]$ we invert the function in the interval, where it is injective, and we get $x=2\sqrt{1-y}$. The interval for $y$ is $[-2;\;1]$

We use the formula for the surface $$S=2\pi \int _{a}^{b}f(y){\sqrt {1+f'(y)^{2}}}\,dy$$ where $f(y)=2\sqrt{1-y}$ and $f'(y)=-\dfrac{1}{\sqrt{1-y}}$, $a=-2;\;b=1$

So we get $$S=2 \pi \int_{-2}^1 2 \sqrt{1-y} \sqrt{1+\left(-\frac{1}{\sqrt{1-y}}\right)^2} \, dy$$ Simplifying $$S=4 \pi \int_{-2}^1 \sqrt{2-y} \, dy=\left[\frac{1}{3} (-2) (2-y)^{3/2}\right]_{-2}^1 =\frac{56 \pi }{3}$$ Hope this helps

Answer 2

HINT

$$ dA= 2 \pi x ds = 2 \pi \sqrt {1+y^{'2} } x\, dx $$

$$ A = \pi \int 2 x \sqrt {1 + x^2/4 }\, dx $$

Let $$x^2= u, 2x dx= du $$

$$ A/\pi= \int_0^{ 12= {(2 \sqrt3})^2} \sqrt {1 + u/4 } \quad du$$

so now you can take it further.