Question

A region R is enclosed by the coordinate axes and the graph of y=k(x-5)^2, k>0. When this region is revolved around the x-axis, the solid formed has a volume of 2500pi cubic units. What is the value of k? Answer: k=2

Answer 1

$y=k(x-5)^2$ touches the x-axis at $x=5$ and has a y-intercept when $x=0$ at

$y=k(0-5)^2=25k$

Using disks, the volume of the revolved region would be given by the integral

$\displaystyle\pi\int_0^5\bigg(k(x-5)^2\bigg)^2\,\mathrm dx=\pi k^2\int_0^5(x-5)^4\,\mathrm dx=\pi k^2\int_{-5}^0x^4\,\mathrm dx$
$\implies 2500\pi=625\pi k^2$
$\implies 4=k^2$
$\implies k=2$

Answer 2

So notice the picture below

thus as LammettHash already pointed out, the area will be over the x-axis and from 0 to 5

$\bf \displaystyle \int\limits_{0}^{5}\ \pi [k(x-5)^2]^2dx\impliedby \textit{using the disk method} \\\\\\ \pi \displaystyle \int\limits_{0}^{5}\ k^2(x-5)^4dx\implies k^2\pi \int\limits_{0}^5 \ (x-5)^4dx \\\\\\ \displaystyle k^2\pi \int\limits_{0}^{5}\ (x^4-20x^3+150x^2-500x+625)dx$

$\bf \\\\\\ k^2\pi \left[ \cfrac{x^5}{5}-5x^4+50x^3-250x^2+625x \right]_0^5 \\\\\\ k^2\pi [625-3125+6250-6250+3125]-k^2\pi [0] \\\\\\ \boxed{625k^2\pi =2500\pi }\implies k^2=\cfrac{2500\pi }{625\pi } \\\\\\ k^2=4\implies k=\pm \sqrt{4}\implies k=\pm 2$