# Volume of revolution: integral calculus

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 $k$.

• I just solved it – David Hoffman Oct 1 '12 at 18:33
• On the other side, is it bounded by the $y$-axis? – André Nicolas Oct 1 '12 at 18:59
• @DavidHoffman, you can answer your own question – Berci Oct 1 '12 at 20:26
• Echoing Berci's comment, it's encouraged on this site to post (and, later, accept) answers to your own questions. Please consider doing so. – Gerry Myerson Oct 2 '12 at 0:30

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.