Question

Take the area enclosed by the curves y = sqrt(x), y = 1, and x = 4. Rotate it around the line x = 5. Find the volume.

While I understand the general problem would call for the shell method

∫[a,b] 2*pi*r*h dx

With r = 5 - x and
h = sqrt(x) - 1

But, what happens say if the question asked if it rotates around x = -6? Would my r equation be something like r = 6 + x?

My teacher always tends to show examples of rotation around the positive x axis, nothing negative yet so I was just curious.

Any help is greatly appreciated!

Answer 1

Yes, if you rotated about x = -6, r=x+6.

As you say, using shells of thickness dx,

v = ∫[1,4] 2π(5-x)(√x-1) dx = 103π/15

You can check your answer by using discs (washers) of thickness dy. In this case,

v = ∫[1,2] π(R^2-r^2) dy
where R=(5-x)=(5-y^2) and r=1
v = ∫[1,2] π((5-y^2)^2-1) dy = 103π/15

Answer 2

Thanks for the help Steve! Greatly appreciated!