Problem:
Find the volume generated when the region bounded by the given curves and lines is revolved about the
x-axis using the disk method. Then find it using the cylindrical shell method and verify that they produce the same result.
\begin{align*}
y &= x^2 \\
x &= 1 \\
x &= 2 \\
\end{align*}
Answer:
Here is a plot of $y = x^2$
https://www.wolframalpha.com/input/?i=plot+x%5E2+x+%3D+0%2C3
With the disk method the general formula for area (when going around the x-axis ) is:
$$ V = \int_a^b \pi r^2 \,\, dx $$
where $V$ is the volume and $r$ is the radius. In this case, we have:
\begin{align*}
V &= \int_1^2 \pi x^2 \,\, dx = \frac{ \pi x^3}{3} \Big|_1^2 \\
V &= \frac{ \pi (2^3)}{3} – \frac{ \pi (1^3)}{3} \\
V &= \frac{ 7\pi}{3}
\end{align*}
Now, I use the cylindrical shell method. Here the general formula for volume is:
$$ V = (\text{circumfrence})(\text{height})(\text{thickness}) $$
In this case, the circumfrence is $2 \pi y$, the height is $x$ and the thickness is $dy$.
\begin{align*}
V &= \int_1^4 2 \pi y x \,\, dy \\
x &= \sqrt{y} \\
V &= \int_1^4 2 \pi y y^{ \frac{1}{2} } \,\, dy = \int_1^4 2 \pi y^{ \frac{3}{2} } \,\, dy\\
V &= \frac{4 \pi y^{ \frac{5}{2} }}{5} \Big|_1^4 = \frac{ 4\pi }{5} \left( 2^5 – 1^5 \right) \\
V &= \frac{ 4\pi }{5} \left( 31 \right) \\
V &= \frac{ 124 \pi }{5}
\end{align*}
Where did I go wrong? I am thinking the problem is with calculating the volume using the cylindrical shell method.
The user Moti suggested that I should use $4-x$ instead of $x$. Here is the updated calculations.
\begin{align*}
V &= \int_1^4 2 \pi y (4 – x) \,\, dy \\
x &= \sqrt{y} \\
V &= \int_1^4 2 \pi y (4 – y^{ \frac{1}{2}) } \,\, dy \\
V &= 2 \pi \int_1^4 y(4 – y^{ \frac{1}{2}) } \,\, dy
\end{align*}
Using an online integral calculator, I find:
$$ \int_1^4 y(4 – y^{ \frac{1}{2}) } \,\, dy = \frac{88}{5} $$
This gives us:
\begin{align*}
V &= 2 \pi \left( \frac{88}{5} \right) \\
V &= \frac{ 176 \pi } {2}
\end{align*}
I believe this answer is still wrong.
Best Answer
This is a region around x axis bound by parabola curve $AB$ rotated around $PQ$ (X-axis). Line $PA$ equation is $x = 1$ and line $QB$ equation is $x = 2$. Volume of the disc repeating your methods -
i) Disc method
\begin{align*} V &= \int_1^2 \pi y^2 \,\, dx \\ &= \pi \int_1^2 x^4\, dx \\ &= \frac{\pi x^5}{5} \Big|_1^2 \\ &= \frac{ 31\pi }{5} \end{align*}
ii) Cylindrical shell method -
So first, the height to the curve has to be taken from line BQ. So, it is $(2-x)$. Second, when you integrate using the bound of the curve between $y = (1, 4)$, you only get volume of the region ARB all around the x-axis. So you need to add volume of the cylinder region APQR or you can integrate between $y = (0, 4)$ and then subtract volume between $y = (0, 1)$.
\begin{align*} V &= \pi \, 1^2 \, (2-1) + \int_1^4 2 \pi y (2-x) \,\, dy \\ &= \pi + 2 \pi \int_1^4 (2y - y^{3/2})\, dy \\ &= \pi + 2 \pi ({y^2} - \frac{2}{5}y^{5/2}) \Big|_1^4 \\ &= \pi + \frac{ 26\pi }{5} \\ &= \frac{ 31\pi }{5} \end{align*}
EDIT: alternatively, complete through integration between $y = 0$ to $4$ and then subtract the region between $y = 0$ to $1$ as mentioned above
\begin{align*} V &= \int_0^4 2 \pi y (2-x) \,\, dy - \int_0^1 2 \pi y (1-x) \,\, dy\\ &= 2 \pi \int_0^4 (2y - y^{3/2})\, dy - 2 \pi \int_0^1 (y - y^{3/2})\, dy\\ &= 2 \pi \int_0^4 2y dy - 2 \pi \int_0^1 y dy - 2 \pi \int_1^4y^{3/2}\, dy \\ &= 2 \pi {y^2}\Big|_0^4 - \pi {y^2} \Big|_0^1 - \frac{4\pi}{5}y^{5/2} \Big|_1^4 \\ &= \frac{ 31\pi }{5} \end{align*}