Video: Finding the Volume of the Solid Generated by the Revolution of the Region under the Curve of a Root Function about the π‘₯-Axis

Find the volume of the solid obtained by rotating the region bounded by the curve 𝑦 = √(π‘₯ + 1) and the lines 𝑦 = 0 and π‘₯ = 4 about the π‘₯-axis.

03:24

Video Transcript

Find the volume of the solid obtained by rotating the region bounded by the curve 𝑦 is equal to the square root of π‘₯ plus one and the lines 𝑦 is equal to zero and π‘₯ is equal to four about the π‘₯-axis.

First, we know it’s that the question is asking us to find the volume which is obtained by rotating a region around the π‘₯-axis. And we’re also told that this region is bounded by the curve 𝑦 is equal to the square root of π‘₯ plus one and the lines 𝑦 equal zero and π‘₯ equals four.

We will start by sketching our bounds to get an idea of where our region is located. We start by sketching the curve 𝑦 is equal to the square root of π‘₯ plus one. We see that this has an π‘₯-intercept when π‘₯ is equal to negative one. Next, we sketch the line 𝑦 is equal to zero, which is the π‘₯-axis, and the line π‘₯ is equal to four. Since we’re told that the region is bounded by these three curves, we can add the region to our sketch.

We now recall that if we have a nonnegative function 𝑓 and if we were to take the region bounded as zero is less than or equal to 𝑦 less than or equal to 𝑓 of π‘₯ and π‘Ž is less than or equal to π‘₯ less than or equal to 𝑏, where π‘₯𝑦 is in our region around the π‘₯-axis. Then we can conclude that the volume of the generated solid is equal to the integral from π‘Ž to 𝑏 of πœ‹ multiplied by our function 𝑓 of π‘₯ squared with respect to π‘₯.

The first thing we notice is that the 𝑦-values in our region are bounded above by the function the square root of π‘₯ plus one and below by zero. So we will just set 𝑓 of π‘₯ to be equal to the square root of π‘₯ plus one. And we notice that the square root of π‘₯ plus one is nonnegative. And so the first prerequisite is true.

Next, we know it’s for our region 𝑅 that, on the left, the values of π‘₯ are just bounded by the point negative one. And on the right, they’re bounded by the line π‘₯ is equal to four. So the values of π‘₯ in our region are bounded below by the line π‘₯ is equal to negative one and above by the line π‘₯ is equal to four.

Therefore, since the question tells us that this region is rotated around the π‘₯-axis, we can conclude that all of the prerequisites are true. Therefore, we can conclude that the volume of the generated solid is equal to the integral from negative one to four of πœ‹ multiplied by the square root of π‘₯ plus one squared with respect to π‘₯.

We’re now in a position to evaluate this integral. The first thing we notice is that πœ‹ is just a constant. So we can just take this πœ‹ out of our integral. Next, if we square the square root of π‘₯ plus one, then we just get π‘₯ plus one. This gives us πœ‹ multiplied by the integral from negative one to four of π‘₯ plus one with respect to π‘₯.

The next step is to calculate the antiderivatives of π‘₯ and one, respectively. We have the antiderivative of π‘₯ is π‘₯ squared over two. And the antiderivative of one is just equal to π‘₯. And of course, we ignore our constant of integration because we’re doing a definite integral.

Now all we have to do is evaluate this antiderivative at the limits of our integral. Evaluating the antiderivative at negative one and four gives us πœ‹ multiplied by four squared over two plus four minus negative one squared over two plus negative one. At this point, we can evaluate the exponents and simplify to get an answer of 25πœ‹ divided by two.

Therefore, what we’ve shown is that the volume of the solid obtained by rotating the region bounded by the curve 𝑦 is equal to the square root of π‘₯ plus one and the lines 𝑦 is equal to zero and π‘₯ is equal to four about the π‘₯-axis is equal to 25πœ‹ divided by two.

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