Find the volumes of the solids in Exercises 135 and 136.

Question

135. The solid lies between planes perpendicular to the x-axis at x=-1 and x=1. The cross-sections perpendicular to the x-axis are

a. circles whose diameters stretch from the curve y=-1/√(1+x²) to the curve y=1/√(1+x²).

Accepted Answer

Hello there. Today, we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A solid is bounded by two vertical planes perpendicular to the x axis at x is equal to -2, and x is equal to 2. The cross sections taken perpendicular to the x axis are circular disks. The diameter of each circular cross section extends between the curves. Y is equal to -2 divided by the square root of 4 + x 2. Andy is equal to 2 divided by the square root of 4 + x to the 2. Find the volume of the solid. OK, so it appears for this particular problem, we are ultimately asked to solve for the volume of the solid, provided all the information that is given to us by the prom itself. So now that we know that we're ultimately solving for the volume of the solid, what we must do first is we need to recall, based on our prior knowledge, that we need to express the diameter and radius of each circular cross section at position x. And let's make a note of this, so let's put little stars here around the letter X. So once again, we are expressing the diameter and radius of each circular cross section at position X. So with that said, let's focus on the upper curve first, which we'll just call upper C for short. So focusing on the upper curve first, we could state that Y top or Y subscript top. is equal to 2 divided by the square root of 4 + x to the power of 2, and the lower curve, which we'll call lower c for short, and the lower curve. We could write as Y subscript BOT bottom for short, so Y subscript bottom. Is equal to -2 divided by the square root of or plus x to the power of 2. So therefore that will mean that the diameter will be D X is equal toy top minus y bottom, which will be equal to 2 divided by the square root of 4. Plus x 2 minus 2 divided by the square root of 4 + x 2, which will be all equal to 4 divided by the square root of 4 + x to the power of 2. And now solving for the radius, the radius will be written as R X is equal to D X divided by 2. Which will be equal to 2 divided by the square root of 4 + x to the power of 2. Fantastic. So now, we need to write the area of the circular cross section. So we need to write the equation for the area of the circular cross section. And we'll denote the area as, or rather the function for our area as A of X, and once again A of X, the area which we're solving once again, we're solving for the area of the circular cross section. So A of X will be equal to pi multiplied. By parentheses of r x to the power of 2, which will be equal to pi multiplied by or divided by 4 + x to the power of 2, which will be equal to 4 pi divided by 4 + x to the power of 2. Fantastic. So now, what we need to do is we need to set up the volume integral from, so once again, we're solving for the volume, which we'll call V for short, and focusing our attention on the volume integral from. X is equal to -2 x is equal to 2. So therefore we could write our expression for the volume equation as V is equal to the integral evaluated from -2 to 2 of a of x Dx, which is equal to the integral evaluated from -2 to 2 of. 4 pi divided by 4 plus x to the power of 2 dx. Which will be equal to 4 pi multiplied by the integral evaluated from -2 to 2 of 1 divided by 4 plus x to the power of 2. D X. So now at this stage, we need to evaluate our integral and in order to do that, we need to recall, so let's make a note, we need to recall that the integral of 1 divided by a 2 + X2 DX is equal to 1 divided by a. Multiplied by inverse tangent of x divided by a plus c where c represents some constant variable, so therefore, Let us also note that. A is equal to 2 and because therefore with A is equal to 2, we can go ahead and write that the integral evaluated from -2 to 2 of 1 divided by 4 plus. X to the power of 2, DX will be equal to drum roll, 1 divided by 2. Multiplied by parentheses of. Inverse tangent of x divided by 2. Evaluated from -2 to 2. And this will be equal to 1 divided by 2 multiplied by parentheses of inverse tangent. Of 1, so it's gonna be equal to 1/2 multiplied by parentheses of inverse tangent of 1 minus inverse tangent of -1. Which will be equal to -1/22, or actually it's gonna be equal to 1/2, so this will be equal to 1/2 multiplied by parentheses of pi divided by 4 minus pi divided by 4, which will be equal to when we simplify our answer down to its most simple form, pi divided by 4. So, therefore, without a doubt, our volume of the solid will be equal to, so V, the volume will be equal to 4 pi multiplied by pi divided by 4, which will be simply equal to pi to the power of 2. And that's it. That is our final answer. Hooray, we did it. So going back to the top to recap what we've learned, we could safely say that our final answer, without a doubt, and don't forget our units, our final answer is pi to the power of 2 cubic units. It is once again volume. Answers for the volume are usually in cubic units. And since we're not specifically told that we're using units of centimeters, meters, etc. we need to just make sure we're just talking about units, so we know that it's a unit of measurement. And that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Find the volumes of the solids in Exercises 135 and 136.135. The solid lies between planes perpendicular to the x-axis at x=-1 and x=1. The cross-sections perpendicular to the x-axis area. circles whose diameters stretch from the curve y=-1/√(1+x²) to the curve y=1/√(1+x²).

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Find the volumes of the solids in Exercises 135 and 136.
135. The solid lies between planes perpendicular to the x-axis at x=-1 and x=1. The cross-sections perpendicular to the x-axis are
a. circles whose diameters stretch from the curve y=-1/√(1+x²) to the curve y=1/√(1+x²).