Question

Calculus 2: I have 2 problems I'm stuck on...

Answer 1

Using the method of "disks" or "washers"
You consider a cross-sectional area with ring area \(A_R=\pi\left[(\text{outside ring})^2-(\text{inside ring})^2\right]\) with respect to the y-axis (since you are rotating about the vertical line \(x=0\).

The outside ring radius will be represented by your furthest curve from the y-axis, which is the one \(y=x\). The inside ring radius will be represented by the curve closest to the y-axis, which is \(y=x^3\). But, since you're rotating about the y-axis, you need to express your functions in terms of x.

Thus: \(y=x \implies x=y\)
\(y=x^3 \implies x=y^{1/3}\)

Thus, the cross-sectional ring area is \(A_R=\pi\left[(y^{1/3})^2-(y)^2\right]=\pi(y^{2/3}-y^2)\).

Now, the volume is simply \[V=\int_a^b A_R \, dy\], where \(a\) and \(b\) are the y-values for the endpoints of your figure that you are rotating. From the diagram, you see that the y-value ranges from \(y=0\) to \(y=1\).

Hence, you have \[ V=\int_0^1\pi(y^{2/3}-y^2)\,dy\]

Answer 2

Integrating this should be straightforward. \(\pi\) is just a constant that can go outside the integral, and then you just use the simple power rule to evaluate the integral.