90. Work Let R be the region in the first quadrant bounded by ...

Question

90. Work Let R be the region in the first quadrant bounded by the curve y = √(x⁴ - 4)

and the lines y = 0 and y = 2. Suppose a tank that is full of water has the shape of a solid of revolution obtained by revolving region R about the y-axis. How much work is required to pump all the water to the top of the tank? Assume x and y are in meters.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says let Q be the region of the first quadrant enclosed by Y is equal to the square root of the quantity of X to the 4th minus 16 in quantity, Y equal to 0 and Y equal to 4. This region is revolved around the y axis to form a tank and then completely filled with water. How much work is required to pump all the water up to the top of the tank? Assume that the variables XY are measured in meters, and use whoever water is equal to 1000 kg per meter cubed, and G is equal to 9.8 m per second. So, let's begin by drawing a diagram of what we are given in the problem. So, we're looking at the first quadrant, so we'll draw our X and Y coordinates so we can so that we can see the first quadrant. And we're looking for the region that's enclosed by Y equal to 0, which is our X-axis, the horizontal line, Y equal to 4, and the curve, Y is equal to the square root of the quantity of X at 4th minus 16. So, that is a curve that crosses the x-axis at X equal to 2. And we'll follow a square root function curve up until it reaches Y equal to 4. And so we're gonna take this region and revolve it around the y axis, and then fill it with water. And we need to figure out how much work is required to pump all the water up to the top of the tank. So to do this, we're actually going to look at a small slice of water, and we're gonna make these slices horizontally. And so, the slice is going to be at a height of Y. And so we need to figure out how much work it takes to move this from the height of white up to the top of the tank. So, we're gonna call that work DW since this is just gonna be a small portion of the total work. And this is going to be equal to, and here we call your formula for work, that's going to be the weight, which is going to be our density of water, or water. Multiplied by the acceleration due to gravity, which is G, multiplied by the volume, and here we just have a small slice of volume, so we'll label it as DV, and then this gets multiplied by the distance that we have moved this slice. So since we're moving up to the top of water at Y equal to 4, and we're initially at a height of Y, the distance is just going to be 4 minus Y. So now we need to figure out our small volume of water that we're moving, but we need to find DV. So DV is going to be equal to, and here we need to consider what we're actually doing here. So when we take a thin slice of this water here, we're actually taking a thin slice, which is um in the shape of a circle. Since we're revolving around the y axis, and the radius of circle is just going to be our X variable or our curve. So that means that DV is going to be equal to pi, multiplied by X2d for the area, and we'll need to multiply this by the thickness, which is going to be D Y, since you're going to have some small thickness along the Y direction, so we'll label that as D Y. And so, what we really need to do is write X as a function of Y, since we're ultimately going to be integrating later on with respect to Y since we have D Y in this formula. So to do this, we'll just take our equation for Y and solve it for X. The first step is to square both sides of that equation, and so we'll get Y2 is equal to x 4th minus 16, and then to solve for X we'll add 16 to both sides and then take the 4th root of both sides. In doing so, we'll get that X is equal to the quantity of Y2 + 16 in quantity raised to the 1/4th power. So, substitute this back into our volume formula, this is going to be equal to. Pi multiplied by the quantity of Y2 + 16 in quantity raised to the 12 power. Multiplied by DY. So now we can put everything together for our work formula for this slice. So, we will have DW is equal to, and here we'll substitute in the value for our density of water, which is going to be 1000. Then that gets multiplied by the acceleration due to gravity, which is 9.8. And then this gets multiplied by pi, multiplied by the quantity. Of or minus Y in quantity, multiplied by the square root of the quantity of Y2 + 16 in quantity, multiplied by D Y. And so combining The constants out front, this is going to be equal to 9800 multiplied by pi, multiplied by the quantity of 4 minus Y in quantity, multiplied by the square root with quantity of Y2 + 16 in quantity D Y. So, we now want to find the total work done by moving all of these slices up to the top. So we're going to need to integrate our work here to find the total work. So, our total work, which will be W, that's going to be equal to the integral, and here we're integrating a Y, so that's going to go from 0 to 4. And this is going to be of quantity of 9800, multiplied by pi, multiplied by the quantity of 4 minus Y in quantity, multiplied by the square root of the quantity of Y quad, plus 16 in quantity D Y. And so we want to do is actually break this integral up into two separate integrals by distributing the square root of the quantity of Y2 + 16 into the quantity of or minus Y. In doing so, this is going to be equal to here we'll pull all the constants out front, so we'll have the 9800 multiplied by pi, multiplied by the quantity, and for the first interval we'll have the integral from 0 to 4 of 4 multiplied by the square root of the quantity of Y2 + 16 in quantity D Y. Then we'll have minus the integral from 0 to 4 of Y multiplied by the square root of the quantity of Y2 + 16 in quantityd Y in quantity. So now what we need to do is to integrate each of these integrals separately. So looking at the first integral, we can actually integrate this using our standard formulas for integration. And so recall that the integral of the square root of the quantity of Y2 plus A2 in quantity D Y is equal to. Y divided by 2, multiplied by the square root with quantity of Y2 plus A2 in quantity, plus a 2 divided by 2. Multiplied by the natural log of the absolute value of the quantity of Y plus the square root of the quantity of Y2 plus a squad in quantity. Plus see So, using this formula, that means that the integral from 0 to 4 of the square root of the quantity of Y2 + 16 in quantity, DY is going to be equal to. Y divided by 2 multiplied by the square root of the quantity of Y2 + 16 in quantity, plus 16 divided by 2 multiplied by the natural log of the absolute value of Y plus the square root of the quantity of Y2 plus 16 in quantity. And all of that is going to be evaluated over the interval from 0 to 4. So substituting in the limits of integration, this is going to be equal to. 4 divided by 2 multiplied by the square root of the quantity of 4 quad plus 16 in quantity, plus 8 multiplied by the natural log of the absolute value of 4, plus the square root of the quantity of 4 squad plus 16 in quantity. Minus, or the lower limit, the quantity of 0 divided by 2, multiplied by the square root of the quantity of 0 squared plus 16. In quantity, plus, 8, multiplied by the natural log of the absolute value of the quantity of 0, plus the square root of the quantity of 0 squad plus 16 in quantity. And so simplifying this expression, this is going to be equal to. 8, multiplied by the square root of 2. Plus 8, multiplied by the natural log of the quantity of 1, plus square root of 2 in quantity. Where for the natural log function, we've dropped the absolute value sign since its argument is going to be positive. Now, to integrate the next interval, we're going to use U substitution. So we're gonna set U equal to Y2 + 16, and that means that DU is going to be equal to 2 Y D Y. However, we just have YDY in our integrals, so we'll divide both sides by 2, and we'll get DU divided by 2 is equal to Y D Y. So that means that the interval from 0 to 4 of Y multiplied by the square root of the quantity of Y2 + 16 in quantity. D Y is going to be equal to the integral of, and here we need to change our limits of integration. So when Y is equal to 0, U is going to be equal to 02 + 16, so that's gonna be 16. And when Y is equal to 4 for the upper limit. We'll have U is equal to 42 + 16, which is 32. And then for our integran, we will just have the square root of U multiplied by DU divided by 2. And so performing the integration. This is going to be equal to. One half multiplied by, and here when we integrate the square root of you, we get you to the three halves. How divided by the quantity of 3 divided by 2 in quantity, and this needs to be evaluated at 16 and 32. So, performing the evaluation of the limits of our integral, this is going to be equal to. When we simplify the coefficient out front, we get 1 divided by 3, and then this is multiplied by the quantity of 32 raised to the 3/al power, minus 16 raised to the 3 halves power in quantity, which is equal to 1 divided by 3, multiplied by The quantity of 128 multiplied by the square root of 2. 64 in quantity. So now we need to put everything together with our work function. So our work, W. is going to be equal to the 9800 multiplied by pi. Multiplying the quantity of 4, multiplied by the quantity of 8, multiplied by the square root of 2. Plus 8 multiplied by the natural log of quantity of 1 plus the square root of 2 in quantity, in quantity again. Then we'll have minus. The quantity of 1 divided by 3, multiplying the quantity of 128, multiplied by the square root of 2 minus 64 in quantity. So now we just need to simplify this expression. And so we can factor out a 32 out of each of these terms. And then collect light terms. In doing so, this is going to be equal to 313,600 high. Multiplied by the quantity of The natural log of the quantity of 1 plus the square root of 2 in quantity minus the quantity of. The square root of 2 minus 2 in quantity divided by. 3. In quantity. And we need to include units on this, and the units for work is actually joules, since everything was given in meters and kilograms and meters per second squared for our acceleration. So this here is the answer that we're looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Solid of Revolution

Work Done Against Gravity

Integration in Calculus

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