Question

Let $P\left( 4,k \right)$ be any point on the line $y=6-x$. If the vertical segment $PQ$ is rotated about $y$-axis, the volume of the resulting cylinder is

Answer 1

Hint: In this question we have been given with a point $P\left( 4,k \right)$ which is on the line $y=6-x$. We will use the equation of the line and the coordinate of the point $P$ to find the value of the $y$ coordinate. We will then use the formula of volume of a cylinder which is given by $V=\pi {{r}^{2}}h$ and substitute the values of the radius and height to get the required volume.

Complete step by step solution:
We know that point $P\left( 4,k \right)$ lies on the line $y=6-x$.
From point $P\left( 4,k \right)$ we can see that the $x$ coordinate is given to us and we need to find the $y$ coordinate therefore, we will substitute $x=4$ in the equation of the line to get the value of $y$ which is the $y$ coordinate.
On substituting $x=4$ in $y=6-x$, we get:
$\Rightarrow y=6-4$
On simplifying, we get:
$\Rightarrow y=2$
Therefore, we have the $y$ coordinate as $2$.
The point $P$ becomes $P\left( 4,2 \right)$.
Now we know that the vertical segment $PQ$ is rotated therefore we have the height of the cylinder as the value of the $y$ coordinate and the radius of the cylinder as the value of the $x$ coordinate.
Therefore, we can write:
$r=4$
$h=2$
Now the formula for the volume of a cylinder is $V=\pi {{r}^{2}}h$. On substituting the value of $r$ and $h$, we get:
$\Rightarrow V=\pi {{\left( 4 \right)}^{2}}2$
On simplifying, we get:
$\Rightarrow V=32\pi $, which is the required solution.

Note: In this question we have been told that the vertical segment $PQ$ is rotated. If the question had mentioned that the horizontal segment formed at the intersection of $y$ axis and the point $P$, the value of the height and the radius would have been inverted as $4$ and $2$ respectively, hence the volume would have been different.