Video: Applications of Integration: Volumes by Slicing

Use the slicing method to find the volume of the solid whose base is the region bounded by the lines π‘₯ + 5𝑦 = 5, π‘₯ = 0, and 𝑦 = 0 if the cross sections taken perpendicular to the π‘₯-axis are semicircles.

06:09

Video Transcript

Use the slicing method to find the volume of the solid whose base is the region bounded by the lines. π‘₯ plus five 𝑦 is equal to five, π‘₯ is equal to zero, and 𝑦 is equal to zero. If the cross sections taken perpendicular to the π‘₯-axis are semicircles.

The question is asking us to use the slicing method to find the volume of this solid. And we recall we can use the slicing method in the following four steps. First, we need to determine the region which represents the base of our solid. And We can do this by sketching the information given to us in the question.

Next, we need to determine the shape and orientation of the cross sections of our solid. And again, we can do this by sketching the information given to us in the question. Third, we need to find the area of our cross sections. And if our cross sections are taken perpendicular to the π‘₯-axis, we want our area in terms of π‘₯. If our cross sections are perpendicular to the 𝑦-axis, we want our area in terms of 𝑦. And if our cross sections are perpendicular to the 𝑧-axis, we want our area in terms of 𝑧. Finally, we just integrate our formula for the area over an appropriate interval. And then this gives us the volume of our solid.

Let’s start by finding the base of our solid. The question tells us the base of our solid is bound by the lines π‘₯ plus five 𝑦 is equal to five, π‘₯ is equal to zero, and 𝑦 is equal to zero. Since we’re told these are lines, this tells us that our base lies within the π‘₯𝑦-plane. So let’s sketch a graph of our base.

We know the lines π‘₯ is equal to zero and 𝑦 is equal to zero coincide with our axes. And we can rearrange the straight line π‘₯ plus five 𝑦 is equal to five to give us 𝑦 is equal to negative π‘₯ over five plus one. And since this is a standard form for a straight line, we know that the slope of this straight line is negative one-fifth. And the 𝑦-intercept is one.

So we can sketch a graph of π‘₯ plus five 𝑦 is equal to five. We know it intercepts the 𝑦-axis when 𝑦 is equal to one. And we can also find the π‘₯-intercept. This will be when π‘₯ is equal to five. Since the π‘₯-intercept will be when 𝑦 is equal to zero. Now since our base is bounded by the lines π‘₯ is equal to zero, 𝑦 is equal to zero, and π‘₯ plus five 𝑦 is equal to five, we can see that we found the sketch of our base. In fact, it’s a right-angled triangle with height one and width five. So we found that our base is a right-angled triangle.

Next, we need to determine the shape of our cross sections. And the question tells us that any cross section taken perpendicular to the π‘₯-axis will be a semicircle. So let’s use this information to sketch our cross sections.

We’ll start by sketching our base on a set of three-dimensional axes. Next, we’re told that the cross sections taken perpendicular to the π‘₯-axis are semicircles. So we’ll draw a line which is perpendicular to the π‘₯-axis. And this cross section will be a semicircle. In fact, we can also see the diameter of our semicircle will be the height of our perpendicular line.

So we now have that our cross sections are semicircles perpendicular to the π‘₯-axis with a diameter equal to the height of our perpendicular line. We now need to calculate the area of our cross section. And remember, since we’re taking our cross sections perpendicular to the π‘₯-axis, we want our formula in terms of π‘₯. And we recall the area of a circle with radius π‘Ÿ is πœ‹π‘Ÿ squared. So the area of a semicircle will be πœ‹π‘Ÿ squared all divided by two.

Now to calculate the area of our cross sections, we’ll go back to the sketch of our base. And we know the diameter of a cross section taken at π‘₯ will be the length of this vertical line. And the length of this vertical line will just be its 𝑦-coordinate. And we found that its 𝑦-coordinate is equal to negative π‘₯ divided by five plus one. So our cross sections have a diameter of negative π‘₯ divided by five plus one. And if we divide this by two, we’ll find the radius of our cross section.

So we can calculate the area of our cross section by using a half πœ‹π‘Ÿ squared. This gives us πœ‹ over two multiplied by a half multiplied by negative π‘₯ over five plus one squared. Then distributing our exponent over our parentheses and simplifying, we have πœ‹ over eight multiplied by π‘₯ squared over 25 minus two π‘₯ over five plus one.

So we’ve now found a formula for the area of our cross section. And since our cross section is given perpendicular to the π‘₯-axis, we’ve written our area in terms of π‘₯. Finally, we need to integrate our area over an appropriate interval to find the volume of our solid.

We can see from our sketch that the values of π‘₯ vary from zero to five. This tells us we can calculate the volume of our solid as the integral from zero to five of the area of our cross sections perpendicular to the π‘₯-axis with respect to π‘₯. So by using the formula we found for the areas of our cross section, we have that our volume is equal to the integral from zero to five of πœ‹ over eight times π‘₯ squared over 25 minus two π‘₯ over five plus one with respect to π‘₯.

First, we’re going to take the constant coefficient of πœ‹ over eight outside of our integral. Then we’re going to integrate each term by using the power rule for integration. This gives us πœ‹ over eight multiplied by π‘₯ cubed over 75 minus π‘₯ squared over five plus π‘₯ evaluated at the limits of our integral. π‘₯ is equal to zero and π‘₯ is equal to five.

Evaluating at the upper limit of our integral gives us five cubed over 75 minus five squared over five plus five. We then need to subtract the evaluation at the lower limit of our integral. However, we see that all three terms share a factor of π‘₯. So when we evaluate this at π‘₯ is equal to zero, we will get zero. Evaluating this expression gives us πœ‹ over eight multiplied by five over three, which simplifies to give us five πœ‹ over 24.

Therefore, by using the slicing method, we found the volume of the solid whose base is the region bounded by the lines. π‘₯ plus five 𝑦 equals five, π‘₯ is equal to zero, and 𝑦 is equal to zero. With cross sections perpendicular to the π‘₯-axis are semicircles is equal to five πœ‹ divided by 24.

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