The question is the following:
Find the surface area given by the revolution of the curve $y = 1 – x$ with $0 \leq x \leq 1$ around the y axis
In order to do that I made the following "substitution":
$x = 1 – y$ and $0 \leq y \leq 1$ so I can have an interval for $y$
Then we have:
$$x = f(y)\cos\theta \\ y = y \\ z = f(y)\sin\theta$$
Since $f(y) = 1-y$, then:
$$x = (1-y)\cos\theta \\y = y \\ z = (1-y)\sin\theta$$
Cross product between $r_y$ and $r_{\theta}$:
$$||r_y\cdot r_\theta||= (1-y)\sqrt{2}$$
Finally, the integral:
$$\int\int_D||r_y\cdot r_\theta||\text{dA} = \int^{2\pi}_0\int^1_0(1-y)\sqrt{2}\,\,\,dyd\theta=\pi \sqrt{2}$$
Is it okay to perform the "substitution" I did (assuming the interval of $y$ based on the function and the given interval)? I'm also not sure if this is 100% right, can someone check this please?
Best Answer
Yes you eventually get to the answer but you did not need a substitution.
The given curve is $y = x - 1$. The arc length element $dl = \sqrt{1 + \left(\frac{dx}{dy}\right)^2} ~ dy = \sqrt2 ~ dy$
Also, $0 \leq x \leq 1 \implies -1 \leq y \leq 0$. As we are rotating around y-axis, the radius of rotation at a given $y$ is $x = 1 + y$. So the integral to find surface area is,
$S = \displaystyle \sqrt2 \int_{-1}^0 2 \pi (1 + y) ~ dy = \pi \sqrt2$