1. Rotating two curves

1) Rotate the graph 4x^5 about the line x = 1 and calculate the volume of the generated figure.

I assumed that the radius $\displaystyle r = 1 - 4x^5$ and then integrated from 0 to 1 and was going to double it

$\displaystyle A = \pi (1-4x^5)^2$
$\displaystyle A = \pi (1-8x^5+16x^{10}$
$\displaystyle \int^1_0 A(x)dx = x-\frac{4}{3}x^6 + \frac{16x^{11}}{11}|^1_0$

Which gives you $\displaystyle 1-\frac{4}{3}+\frac{1}{11}$

2) $\displaystyle y = 5x, y = 5\sqrt{x}$ Rotated about y = 5. Find the volume

Washer method:

$\displaystyle A(x) = \pi (5-5x)^2-\pi (5-\sqrt{x})^2$
$\displaystyle \pi \int^1_0 (25-50x+25x^2)-(25-50\sqrt{x}-25x)$
$\displaystyle \pi \int^1_0 -75x+25x^2+50\sqrt{x}$

But that evaluates to 0, which doesn't make sense (and is also wrong).

Where did I go wrong on these?

2. If you are rotating R1 about x= 1 use the shell method

Height is 4*x^5 and the distance from x to 1 is 1-x

you have (2pi)integral[(1-x)4x^5dx]

3. Originally Posted by Open that Hampster!
1) Rotate the graph 4x^5 about the line x = 1 and calculate the volume of the generated figure.

2) $\displaystyle y = 5x, y = 5\sqrt{x}$ Rotated about y = 5. Find the volume
(1) I assume the x-axis is the lower boundary for the region ...

using cylindrical shells ...

$\displaystyle V = 2\pi \int_0^1 (1-x)(4x^5) \, dx$

using washers ...

$\displaystyle y = 4x^5$ ... $\displaystyle x = \left(\frac{y}{4}\right)^{\frac{1}{5}}$

$\displaystyle V = \pi \int_0^4 \left[1 - \left(\frac{y}{4}\right)^{\frac{1}{5}}\right]^2 \, dy$

(2) using washers ...

$\displaystyle R(x) = 5 - 5x$

$\displaystyle r(x) = 5 - 5\sqrt{x}$

$\displaystyle V = \pi \int_0^1 (5-5x)^2 - (5 - 5\sqrt{x})^2 \, dx$

4. We never learned the shell method, so I'll give that a go.

As for number 2, that's what I did skeeter, but I didn't get the right answer. Was it an Algebraic mistake?