As part of my Calculus 2 class, I have created a practice exam to help my students study. In doing so, I felt that providing the questions and the solutions in a blog series would be helpful not just to my students, but also to anyone that may be taking, or teaching, calculus 2. I hope you find the following solutions helpful in your studies.

You can also view me presenting the solution on YouTube at

**Let \(R\) be the region bounded by \(f(x)=x^{2}\), \(x=1\) and \(y=0\). Find the volume of the object obtained by revolving \(R\) about the \(y\)-axis.**

We begin by graphing these and finding the corresponding region \(R\).

If we then revolve this around the \(y\)-axis, we will get the following picture.

Since we want to find the total volume, we will have to cut out cross sections. When cutting cross sections, we now have two options. One of the options will be cutting out washers, the other will be to cut out shells. We will go through both options.

**Washers**

Using these as cross sections, we find that the volume is the area of the face of the cross sections multiplied by the depth. Since the face is a big circle with a smaller circle cut out, we can find the volume of the cross section to be

\begin{align*}

V_{cs}&=A*d \\

&=(A_{b}-A_{s})d \\

&=(\pi R^{2}-\pi r^{2})d.

\end{align*}

We now note that the depth is moving up and down, and therefore will be the change in the \(y\)-direction. We also note that the bigger radius is \(x\) when \(x=1\) and the smaller radius is \(x\) when \(y=x^{2}\). Solving for \(x\) for the smaller radius, we find that the smaller radius is \(\sqrt{y}\). We, therefore, get that

\begin{align*}

V_{cs}&=(\pi (1)^{2}-\pi (\sqrt{y})^{2})\Delta y \\

&=\pi(1-y)\Delta y.

\end{align*}

We can now find that the total volume will be

\begin{align*}

V_{T}=\int_{a}^{b}\pi (1-y) dy

\end{align*}

The smallest \(y\) we need is given in the problem as \(y=0\). For the largest \(y\), we find the intersection of \(y=x^{2}\) and \(x=1\). Therefore \(y=1^{2}=1\). We then have that

\begin{align*}

V_{T}&=\int_{0}^{1}\pi (1-y) dy \\

&=\pi(y-\frac{y}{2})|_{0}^{1} \\

&=\pi(1-\frac{1}{2}-0)\\

&=\frac{\pi}{2}.

\end{align*}

**Washers**

We find that the volume is the area of the face of the cross sections multiplied by the depth. If we cut the face of the cross section and pull it straight, we will end up with a rectangle. The height of this rectangle will be the \(f(x)=x^{2}\) for the given \(x\). The length of the rectangle will be the circumference of a circle with radius \(x\), that is, \(2\pi x\). Furthermore, the depth of the shell will be the change in the \(x\) direction. We can now find the volume of the cross section to be

\begin{align*}

V_{cs}&=A*d \\

&=2 \pi x * x^{2}\Delta x

\end{align*}

We can now find that the total volume will be

\begin{align*}

V_{T}=\int_{a}^{b}2 \pi x^{3} dy

\end{align*}

The smallest \(x\) we need corresponds to have a circle of radius 0, that is \(x=0\). For the largest \(x\), we note that the bound is given by the problem as \(x=1\). Therefore, we have that

\begin{align*}

V_{T}&=\int_{0}^{1}2 \pi x^{3} dyx \\

&=2\pi\frac{x^{4}}{4}|_{0}^{1} \\

&=\frac{\pi}{2}-0\\

&=\frac{\pi}{2}.

\end{align*}

**Other Resources**

I have it linked above, but you can find the practice exam here. You can find the other solutions to the practice exam in the other posts available here.

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