The portal has been deactivated. Please contact your portal admin.

Question Video: Setting up the Integral for the Volume of the Solid Generated by the Revolution of the Region Bounded by an Exponential Function Curve about a Line Parallel to the π‘₯-Axis Mathematics

Set up an integral for the volume of the solid obtained by rotating the region bounded by the curve 𝑦 = 𝑒^(βˆ’π‘₯Β²) and the lines 𝑦 = 0, π‘₯ = βˆ’5, and π‘₯ = 5 about 𝑦 = βˆ’5.

04:07

Video Transcript

Set up an integral for the volume of the solid obtained by rotating the region bounded by the curve 𝑦 is equal to 𝑒 raised to the power negative π‘₯ squared and the lines 𝑦 is zero, π‘₯ is equal to negative five, and π‘₯ is equal to five about 𝑦 is equal to negative five.

In this question, we’re given a region bounded by the curve 𝑦 is equal to 𝑒 raised to the power negative π‘₯ squared, the line 𝑦 is equal to zero, and the two vertical lines π‘₯ is equal to positive five and π‘₯ is equal to negative five. And we want to set up an integral for the volume of the solid obtained by rotating this region about the line 𝑦 is equal to negative five. This means that the line 𝑦 is equal to negative five is our axis of rotation.

It can be helpful to try and sketch the solid we’d like to find the volume of, but note that our two sketches have different scales. Now, the solid formed by rotating the given region has washers, that is, hollowed disks, as vertical cross sections. The cross sections have both an inner and an outer radius. Let’s call the outer radius 𝑅 sub 𝑂 and the inner radius 𝑅 sub 𝐼. And now recalling that the area of a circle with radius π‘Ÿ is equal to πœ‹π‘Ÿ squared, then the area of our cross-sectional disks 𝐴 sub 𝐷 is given by πœ‹ times 𝑅 outer squared minus πœ‹ times 𝑅 inner squared. And taking πœ‹ outside some parentheses, that’s πœ‹ multiplied by 𝑅 outer squared minus 𝑅 inner squared.

Now, if we consider the outer radius of one of our disks on our original diagram, the outer radius is the distance from the center of rotation, that’s the line 𝑦 is equal to negative five, and the function 𝑦 is equal to 𝑒 raised to the power negative π‘₯ squared, that is, 𝑒 raised to the power negative π‘₯ squared plus five. So our outer radius varies depending on the value of π‘₯. And remember, π‘₯ ranges from negative five to plus five. If we next consider the inner radius 𝑅 𝐼, that’s the distance between the line 𝑦 is equal to zero and 𝑦 is equal to negative five, and that’s five.

Now, to find the volume of this solid, we take the infinite sum of the areas of all our cross-sectional disks between π‘₯ is negative five and π‘₯ is plus five. That is, the volume is the integral between negative five and five of the area of the disk with respect to π‘₯. And our integral is with respect to π‘₯, since the axis of rotation is parallel to the π‘₯-axis. And so, for our volume, we’re integrating πœ‹ multiplied by the outer radius squared minus the inner radius squared with respect to π‘₯ between negative five and five.

We’re asked to set up an integral for this volume. So all we need to do now is to work out what the outer radius squared minus the inner radius squared is. And the outer radius squared is 𝑒 raised to the power negative π‘₯ squared plus five squared, that is, 𝑒 raised to the power negative two π‘₯ squared plus 10 multiplied by 𝑒 raised to the power negative π‘₯ squared plus 25. The inner radius squared is five squared, and that’s 25. And so we have the outer radius squared minus the inner radius squared is 𝑒 raised to the power negative two π‘₯ squared plus 10𝑒 raised to the power negative π‘₯ squared plus 25 minus 25. And since 25 minus 25 is zero, we have the outer radius squared minus the inner radius squared is equal to 𝑒 raised to the power negative two π‘₯ squared plus 10𝑒 raised to the power negative π‘₯ squared.

Now, making some space, we can substitute this into our integral. This gives us that our volume is the integral between negative five and five of πœ‹ multiplied by 𝑒 raised to the power negative two π‘₯ squared plus 10𝑒 raised to the power negative π‘₯ squared with respect to π‘₯. And since we can take the factor of πœ‹ outside our integral, we have that the volume of the solid obtained by rotating the region bounded by the curve 𝑦 is equal to 𝑒 raised to the power negative π‘₯ squared and the lines 𝑦 is equal to zero, π‘₯ is negative five, and π‘₯ is positive five about the line 𝑦 is equal to negative five is given by πœ‹ multiplied by the integral between π‘₯ is negative five and π‘₯ is positive five of 𝑒 raised to the power negative two π‘₯ squared plus 10 multiplied by 𝑒 raised to the power negative π‘₯ squared with respect to π‘₯.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.