Video: Finding the Volume of a Solid Whose Base Is Bounded by a Function of π‘₯ in the First Quadrant and Whose Semicircular Cross Sections Are Perpendicular to the π‘₯-Axis

The base of a solid is the first quadrant region bounded by 𝑦 = ∜6 βˆ’ 2π‘₯, and each cross section perpendicular to the π‘₯-axis is a semicircle with a diameter in the π‘₯𝑦-plane. What is the volume of the solid?

03:50

Video Transcript

The base of a solid is the first quadrant region bounded by 𝑦 is equal to the fourth root of six minus two π‘₯, and each cross section perpendicular to the π‘₯-axis is a semicircle with a diameter in the π‘₯𝑦-plane. What is the volume of the solid?

We’re asked to find the volume of a solid whose base is bounded in the first quadrant region by the function 𝑦 is equal to the fourth root of six minus two π‘₯. Cross sections of the solid are semicircles perpendicular to the π‘₯-axis. And these semicircles have their diameter in the π‘₯𝑦-plane. To find the volume of a solid defined in this way, we find the integral of the area of the cross sections of the solid.

The limits of integration are defined by the boundary function of the base of the solid. So, let’s begin by sketching this boundary function in the π‘₯𝑦-plane. Our base is in the first quadrant. And for the limits of integration, we need to know where the function crosses the π‘₯-axis. That’s when 𝑦 is equal to zero. So, zero is equal to the fourth root of six minus two π‘₯. Remembering that the fourth root means that the exponent is one over four, if we take the fourth power on both sides, we have zero equal to six minus two π‘₯. And solving this for π‘₯ gives us π‘₯ equal to three. Our function, therefore, meets the π‘₯-axis at π‘₯ equal to three. And the region bounded by this function is the base of our solid. Our limits of integration will, therefore, be zero and three.

The cross sections of our solid are defined as semicircles perpendicular to the π‘₯-axis with a diameter in the π‘₯𝑦-plane. Let’s try and sketch how this might look in three dimensions. The diameter of each semicircular cross section has length 𝑦. And since the diameter is twice the radius, the radius of each semicircular cross section is 𝑦 over two. Remember that, to find the volume of our solid, we’re going to find the integral of the area of the cross sections. So, we need to know the area of the cross sections.

Remember that the area of a circle is πœ‹π‘Ÿ squared. So, the area of a semicircle is πœ‹π‘Ÿ squared over two. So, in our case, our area, which is a function of the radius, is πœ‹π‘Ÿ squared over two, where π‘Ÿ is equal to 𝑦 over two. So that our area is πœ‹ by two times 𝑦 over two squared. And that’s πœ‹ by two times 𝑦 squared over four, which is πœ‹ over eight 𝑦 squared. But remember that 𝑦 is our boundary function. It’s defined as the fourth root of six minus two π‘₯. And that’s equal to six minus two π‘₯ to the power one over four.

For our area function, we want 𝑦 squared. And 𝑦 squared is six minus two π‘₯ to the power one over four all squared. That’s six minus two π‘₯ to the power two over four, which is six minus two π‘₯ to the power of a half. And that’s the square root of six minus two π‘₯. So, now, we have our area as a function of π‘₯. And that’s πœ‹ by eight times the square root of six minus two π‘₯. Our volume is, therefore, the integral of πœ‹ over eight times the square root of six minus two π‘₯ with respect to π‘₯.

Since πœ‹ by eight is a scalar, we can put this in front of our integral. And now, what we need to do is find the limits of integration. Remember that the base of our solid is bounded by the function 𝑦 is the fourth root of six minus two π‘₯ in the first quadrant. So, we’re going to be integrating the area of the semicircular cross sections along the π‘₯-axis from zero to three. Our limits of integration are, therefore, zero and three.

So, the volume of the solid whose base is bounded in the first quadrant by the function 𝑦 is the fourth root of six minus two π‘₯ and whose cross sections are semicircles perpendicular to the π‘₯-axis with diameter in the π‘₯𝑦-plane. Is πœ‹ by eight times the integral between zero and three of the square root of six minus two π‘₯ with respect to π‘₯. Note that we can evaluate this integral by using the substitution 𝑒 is equal to six minus two π‘₯, which gives us an approximate volume of 1.475 units cubed.

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