(Integration) Find volume using disk method when rotated about the X-axis

[DarK1]

x = 2y - y²
x = y

 

[MarkFL]

Re: finding VOLUME using integration (DISK METHOD)

The first thing I would do is sketch the region to be revolved:

View attachment 7885

Now, what is the axis of rotation?

 

[MarkFL]

I didn't realize you had posted the same problem twice, and in the second one I saw, gave the axis of rotation. I have since merged the two threads.

Okay, so we have the axis of rotation as the $x$-axis. Can you identify the following:

• The outer radius
• The inner radius
• The limits of integration

 

[MarkFL]

To follow up, the washer method gives us:

$$V=\pi\int_{x_1}^{x^2} R^2-r^2\,dx$$

The volume of an arbitrary washer is:

$$dV=\pi\left(R^2-r^2\right)\,dx$$

The outer radius is the line $y=x$, and so:

$$R^2=x^2$$

The inner radius is the parabolic curve, but we are given this curve in the form $x(y)$, and we need it in the form $y(x)$. So, we may write it in the form:

$$y^2-2y+x=0$$

We see the axis of symmetry is $y=-\dfrac{-2}{2(1)}=1$, and so we want the lower half of the curve.

Apply the quadratic formula, and discard the larger root:

$$y=\frac{2-2\sqrt{1-x}}{2}=1-\sqrt{1-x}$$

Thus:

$$r^2=(1-\sqrt{1-x})^2=1-2\sqrt{1-x}+1-x=2-2\sqrt{1-x}-x$$

Hence:

$$dV=\pi\left(x^2+x+2\sqrt{1-x}-2\right)\,dx$$

To determine the limits of integration, we equate the two curves:

$$x=1-\sqrt{1-x}$$

Let $u=1-x$, and we have:

$$u=\sqrt{u}\implies u\in\{0,1\}\implies x\in\{0,1\}$$

Hence, adding the slices by integration, we obtain:

$$V=\pi\int_0^1 x^2+x+2\sqrt{1-x}-2\,dx=\frac{\pi}{6}$$

Let's check our result using the shell method:

$$V=2\pi\int_0^1 y\left(\left(2y-y^2\right)-y\right)\,dy=2\pi\int_0^1 y^2-y^3\,dy=\frac{\pi}{6}\quad\checkmark$$