The volume of the solid formed when the region bounded by the curve $y = e^x – k$ , the $x$–axis and the line $x = \ln3$ is rotated about the $x$–axis is $\pi\ln3$ units$^3$ . Find $k$.
Volume of revolution: integral calculus
-
0Echoing Berci's comment, it's encouraged on this site to post (and, later, accept) answers to your own questions. Please consider doing so. – 2012-10-02
1 Answers
The formula for volume of $f(x)$ rotated 360 degrees about the x axis from $a$ to $b$ is:
$V=\pi \int^b_a{[f(x)]^2} dx$
So, the integral w are looking at shows that $f(x)=e^x-k$, that the upper limit, $b$, is $ln(3)$ and that, hypothetically, the lower limit would be where $e^x-k$ intersects with the x-axis. This can be verified to be when $e^x-k=0$:
$e^x-k=0$ $e^x=k$ $x=ln(k)$
We know that $k$ cannot be zero since $ln(0)$ is undefined and that $ln(k)$, which is or lower boundary cannot be our upper boundary, $b$. W have a interval for k which is $k \in (0,3)$
Our integral is therefore defined as:
$V=\pi \int^{ln(3)}_{ln(k)}{(e^x-k)^2} dx$
We can evaluate this to:
$V=\pi \left.\left[\frac{e^{2x}}{2}-2ke^x+xk^2 \right] \right|^{ln(3)}_{ln(k)}$
When we compute this, we can simplify this to:
$V=\pi \left[\left(\frac{3}{2}+ln \left(\frac{3}{k}\right)\right)k^2-6k+4.5\right]$
Since we know the volume already $\left(V=\pi ln(3) units^3 \right)$, we can make an equality:
$\pi ln(3)=\pi \left[\left(\frac{3}{2}+ln \left(\frac{3}{k}\right)\right)k^2-6k+4.5\right]$
Eliminate $\pi$,
$ln(3)=\left[\left(\frac{3}{2}+ln \left(\frac{3}{k}\right)\right)k^2-6k+4.5\right]$
And in order for the equality to hold true, $k=1$.
So therefore, in order for:
$\pi ln(3)=\pi \int^{ln(3)}_{ln(k)}{(e^x-k)^2} dx$
To hold true, k must be equal to 1.