A region R is enclosed by the coordinate axes and the graph of y=k(x-5)^2, k0. 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

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

zehanz · Answer

To get started, here is a graph of the situation (I took k=0.1, but it can be any positive value)

anonymous · Answer

okay

anonymous · Answer

do i need to use integrals?

zehanz · Answer

Yes, you do. Do you know the integral for a solid of revolution (about the x-axis)?

anonymous · Answer

no:/

zehanz · Answer

If you have a function f(x) this formula is

$V=\pi\int_{a}^{b}(f(x)^2 dx$

zehanz · Answer

I think you need to know this formula to be able to solve the problem.

Plug in your function:

$2500\pi=\int_{0}^{5}(k(x-5)^2)^2dx$

zehanz · Answer

Oops, forgot the pi on the RHS! It cancels, anyway :D

zehanz · Answer

Could you try to do some work on the integrand before integrating?

anonymous · Answer

foil the (x-5)^2?

zehanz · Answer

That would give you a lot of extra work!
Just keep the brackets, x-5 only means that the graph has shifted 5 units to the right, and has no other meaning. You can keep it while integrating, but first you will have to square k and (x-5)²...

zehanz · Answer

So we now have:

$2500\pi=\pi\int_{0}^{5}k^2(x-5)^4dx$

zehanz · Answer

Simplify by canceling the pi and putting k² before the integral sign (it is a constant):

$2500=k^2\int_{0}^{5}(x-5)^4dx$

anonymous · Answer

okay

zehanz · Answer

So now you have to integrate $(x-5)^4$. 

Do you know how to integrate $x^n$ ?

anonymous · Answer

i am confused about what to do with the exponent

zehanz · Answer

Use the rule $\int x^ndx=\dfrac{1}{n+1}x^{n+1}$

zehanz · Answer

Just remember to use x-5 instead of x, everything is shifted 5 units to the right.

anonymous · Answer

what about the k^2?

zehanz · Answer

Leave it for now, we'll handle it later.

anonymous · Answer

so (x-5)^4 will be (x^2/2 - 5x)^4?

zehanz · Answer

No, it will be $\dfrac{1}{5}(x-5)^5$

zehanz · Answer

Remember the rule and set n = 4.

anonymous · Answer

whats dfrac?

zehanz · Answer

Don't you see a nice fraction? It is supposed to be like this:

anonymous · Answer

ooh okay got it

anonymous · Answer

now what do i do?

zehanz · Answer

Solve this:

zehanz · Answer

It will give you k=2.

Hope you'll succeed. Sorry, have to go now...

anonymous · Answer

got it thanks a lot!:)