Use the general slicing method to find the volume of the follo...

Question

Use the general slicing method to find the volume of the following solids.

The solid whose base is the region bounded by the curves y=x^2 and y=2−x^2, and whose cross sections through the solid perpendicular to the x-axis are squares

fig

Accepted Answer

Hello. In this video, we are going to let the base of a solid be the region R bounded by the curves Y is equal to 2 X2 and Y is equal to 8 minus 2X2. If cross sections perpendicular to the X axis are squares, what is the volume of the solid? We want to use the general slicing method. So, when it means to use the general slicing method means that we want to use the the disc method, but we want to keep in mind that the cross sections that we are cutting now are in the shapes of squares. Now, the area of a square is defined as one of the sides squared, but in this case here, in order to determine the length of the side, the side is going to be defined as the upper curve of the region. Minus the lower curve. Now, in this case here, that is going to be 8 minus 2 X2 minus 2 X2, and this will further simplify to be 8 minus 4 X squad. So what this means is that for one cross section, one side length of the square is going to be 8 minus 4X2d, but the area of the cross section itself is going to be this quantity squared. So the area for the cross section is going to be defined as 8 minus 4 X 2 quantity squared. Now what we want to do is we want to go ahead and determine the volume of this or the area of this region. Now, in order to determine the area, we're going to need to use an integral. This integral is going to be the form of some boundary A to B, of our area function integrated with respect to X. Now, in this case here, we just defined the integrand. It is going to be 8 minus 4 X2 quantity squared. But what we need to do now is we need to define the bounds of integration. Now, the bounds of integration are going to be defined at the points of intersection between the two curves. In order to solve for these points of intersection, we can set the two curves equal to each other and solve for X, since we are integrating with respect to X. That is going to give us 8 minus 2 X squared is equal to 2 X. 2 X2. If we add 2 X2 to both sides of the equation, we get 4 X 2 is equal to 8. Divided by 4 gives us X2 equal to 2, and this is going to leave us with X's equal to positive and negative square root of 2. So that is going to be our bounds of integration. Now, if we plug this into the integral, we can write an integral that will give us the area for the region R. That integral is going to be the integral from negative square root of 2 to square root of 2 of our area function that we defined earlier, which is going to be the quantity ain't minus 4 X2 quantity squared DX. Now, what we want to do is we want to solve this integral. Now notice that we have opposite symmetric symmetric bounds. Because our function is an even function, we can further simplify the bounds and make this equivalent to two multiplied from 0 to square root of 2 of the integran A minus 4 X 2 quantity squared X. We can do this because we are working with symmetric bounds over an even function. Now, what we're going to do is we're going to algebraically simplify the integrand, and in order to do this, we are going to expand the integrand. We can rewrite the integral to be the integral or to multiply by the integral from 0 to square root of 2 of 64 minus. 64 X square. Plus 16 4th power. DX The next thing we'll need to do is we're going to take the antiderivative of each individual term. That is going to leave us with 2 multiplied by 64 X minus 64 divided by 3 X cubed. Plus 16 divided by 5 x 5th power, and this will be evaluated from 0 to square root of 2. The next thing we're going to do is we're going to use the fundamental theorem of calculus part 2, and plug in the values of the bounds. That is going to leave us with 2 multiplied by the quantity, 64 multiplied by 2 minus 64 multiplied by square root of 2, because our upper bound is square root of 2. Minus 64 divided by 3, multiply by square root of 2 quantity cubed, plus 16 divided by 5 multiplied by square root of 2, rates to the fifth power. Then we will have to subtract the value of the lower bound, but notice that if we plug in zero into every term, we will just go ahead and get a lower value of 0. So at this point, what we're going to do is we are going to simplify and distribute the two. That is going to leave us with 128. By distributing 2 and simplifying, we will have 128 square root of 2. Minus 256 square root of 2 divided by 3 plus 128 square root of 2 divided by 5. And by combining these fractions together, we will go ahead and get a final volume or final area of 1,024 square root of 2 divided by 15 square units, which is going to be the solution to this problem. And with that being said, the overall solution to the problem is going to be C. So, I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

Use the general slicing method to find the volume of the following solids.The solid whose base is the region bounded by the curves y=x^2 and y=2−x^2, and whose cross sections through the solid perpendicular to the x-axis are squares

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Use the general slicing method to find the volume of the following solids.
The solid whose base is the region bounded by the curves y=x^2 and y=2−x^2, and whose cross sections through the solid perpendicular to the x-axis are squares