Blogs, Calculus, Study Help

Volume with rectangular cross sections

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


Suppose that \(R\) is the region bounded by \(f(x)=e^{x^{2}}\), \(y=0\), \(x=0\) and \(x=1\). Furthermore, let this region be expanded to 3-dimensions by letting the cross sectional pieces be rectangles of height \(f(x)\), and depth \(x\). Find the volume of the described object.

We first draw the graphs of \(f(x)=e^{x^{2}}\), \(y=0\), \(x=0\) and \(x=1\). We note that the region bounded by these functions is shaded below.

Next, we will draw the the three dimensional object described in the problem. Here, we pull the region out from the drawing so that the distance coming toward us is the the same as the distance \(x\) is from \(x=0\). Furthermore, the height of this object at any point is the same as the \(y\)-value corresponding to \(f(x)=e^{x^{2}}\).

In order to find the total volume, we will now have to cut out cross sections. We can see that the cross sections will be rectangular solids, therefore, the volume of these cross sections will be
\end{align*}\] where \(h=f(x)\), \(d=x\) and \(w\) is the amount of change in the \(x\) direction, so \(w=\Delta x\). We then get
V_{cs}=xe^{x^{2}}\Delta x.

We can now find the total volume by adding the volume of all the cross sections and getting

Note that the smallest \(x\) we are interested in is \(x=0\) and the largest \(x\) is \(x=1\), since these are the bounds given in the problem. We, therefore, have that

In order to find this definite integral, we will need to find an anti-derivative of \(xe^{x^{2}}\). While integrating this, if we cannot guess the answer, we will have to use a substitution. Here the inside function is \(i=x^{2}\). Therefore \(di=2xdx\), so \(dx=\frac{di}{2x}\). We then get that
\int xe^{x^{2}}dx &=\int xe^{i}\frac{di}{2x} \\
&=\frac{1}{2} \int e^{i} di \\
&=\frac{1}{2}e^{i} + c \\

We can now use this to find that
V_{T}&=\int_{0}^{1}xe^{x^{2}}dx \\
&=\frac{1}{2}e^{x^{2}}|_{0}^{1} \\
&=\frac{1}{2}e^{1}-\frac{1}{2}e^{0} \\


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.

If you found the post, or the YouTube video helpful, please like the post or the video.  Also remember to follow the blog and subscribe to the YouTube channel.  By doing so you will not only be able to find future content, you will also make it easier for other students to find the content when they are studying for Calculus.  Thank you.


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