Given surface area of the cylinder $$x^2 + z^2 = a^2$$ cutted by the cylinder $$y^2 = a(a-x) .$$
Find the area of this surface
Solution:
First, let's visualize the problem. The cylinder $$x^2 + z^2 = a^2$$ is a right circular cylinder centered at the origin with a radius of $a$. The equation $$y^2=a(a−x)$$ represents a shifted paraboloid cylinder. We want to find the surface area of the part of the cylinder that lies inside or on the paraboloid cylinder.
In cylindrical coordinates, we have: $$ x = r \cos(\theta) \\ y = \sin(\theta) \\ z = z $$ Now, we need to find the limits for our triple integral. The limits of integration are determined by the intersection of the two surfaces, which is the boundary of the region we want to find the surface area for.
The equation of the cylinder $$x^2 + z^2 = a^2$$ gives us $$ r^2 = a^2 \Rightarrow r = \pm a $$ The equation of the paraboloid cylinder $$y^2 = a(a-x)$$ can be rewritten in cylindrical coordinates as $$ r^2 \sin^2 (\theta) + ar \cos(\theta) = a^2 $$ Solving for $\theta$, using $r = \pm a$ we get: $$ \theta = 2 \pi n, \, n \in \mathbb{Z} $$ So, our integral becomes: $$ S = \int \limits^{\theta_2}_{\theta_1} \mathrm{d} \theta \int \limits^{r(\theta_2)}_{r(\theta_1)} r \mathrm{d} r $$
Finally, I understand, that I choose wrong way for evaluate integral and get correct value of area.
- Let's take a part of a circle on the $OZ$, i.e. $\theta \in (0; \pi/2)$, and $r =(0, a)$, and so, the area $$ S = 4 \cdot \int_{0}^{\pi/2} \mathrm{d}\theta \int_{0}^{a} \sqrt{1+2r^2} \cdot r \mathrm{d}r = \frac{4 \pi}{2} \left( \frac{(2a^2 +1)^{3/2}}{6} - \frac{2\pi}{6} \right) = \frac{\pi}{3} \left( (2a^2 +1)^{3/2} - 1 \right) $$
First, you need to review what you are integrating. To obtain the area of a surface $S$, we need to parametrize the surface as $f(u,v)$ and use the parallelogram approach, that yields $$\text{Area}(S)=\iint_S1dA=\iint|r_u\times r_v|dudv=\iint\sqrt{1+\left(\frac{\partial f}{\partial u}\right)^2+\left(\frac{\partial f}{\partial v}\right)^2}dudv$$
Moreover, we can choose $(u,v)$ however we like, and usually we use the most convenient representation.
We are interested in the portion of the cylinder $x^2+z^2=a^2$ inside the paraboloid sheet $y^2=a(a-x)$. So, we can parametrize this cylinder section using $x=u,y=v$ as $$f(u,v)=(x,y,z)=(x,y,\pm\sqrt{a^2-x^2})\qquad\text{with}\quad (x,y)\in\mathcal{C}\,,$$ where $\mathcal{C}$ is the projection of this part of the cylinder to the XY plane, i.e. $$\mathcal{C}=\left\{(x,y):|x|\leq a \text{ and }y^2\leq a(a-x) \right\}=\left\{(x,y):|x|\leq a \text{ and }|y|\leq \sqrt{a^2-ax} \right\}$$
We can see that ${\partial z}/{\partial y}=0$. Thanks to symmetry, I will focus on the positive root of $z=\sqrt{a^2-x^2}$, and multiply the resulting area by 2, i.e.
$$\text{Area}(S)=2\iint_{\mathcal{C}}\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2}dxdy$$
Now, \begin{align} \frac{\partial z}{\partial x}=\frac{1}{2}\frac{-2x}{\sqrt{a^2-x^2}}=\frac{-x}{\sqrt{a^2-x^2}} \end{align} and \begin{align} \text{Area}(S)&=2\iint_{\mathcal{C}}\sqrt{1+\frac{x^2}{a^2-x^2}}dxdy=2\iint_{\mathcal{C}}\sqrt{\frac{a^2}{a^2-x^2}}dxdy\\ &=2a\int_{-a}^a\int_{-\sqrt{a^2-ax} }^{\sqrt{a^2-ax}}\frac{1}{\sqrt{a^2-x^2}}\,dydx=2a\int_{-a}^a\frac{1}{\sqrt{a^2-x^2}}\Big(\int_{-\sqrt{a^2-ax}}^{\sqrt{a^2-ax}}1dy\Big)\,dx\\ &=2a\int_{-a}^a\frac{1}{\sqrt{a^2-x^2}}\Big(2\sqrt{a^2-ax}\Big)\,dx=4a\sqrt{a}\int_{-a}^a\sqrt{\frac{a-x}{a^2-x^2}}\,dx\\ &=4a\sqrt{a}\int_{-a}^a\frac{1}{\sqrt{a+x}}\,dx=4a\sqrt{a}\Big(2\sqrt{a+x}\Big|_{x=-a}^{x=a}\Big)=8a^2\sqrt{2}\\ \end{align}