Let R be the region bounded by the following curves. Find the ...

Question

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.

y=x and y=4√x; about the x-axis

Graph showing region bounded by y = x and y = 4√x, shaded area revolved around x-axis forming a solid.

Accepted Answer

Let T be the region bounded by a Y equals the cube root of X and Y equals X in the first quadrant. Find the volume of the solid formed when T is revolved about the X axis. We're given a graph of our two lines in its region. And we have 4 possible answers, being 4 pi divided by 54 pi divided by 1515 pi divided by 4, and 4 pi divided by 3, all in cubic units. Now let's first determine what type. Of equation we need to use. We look at this, we notice that we're rotating around the X axis. We can use the washer method for this. Which is the integral of pi multiplied. By our outer radius. Squared minus our inner radius squared. This goes from A to B. And we will determine our variable of integration in a moment. Now solve this We first need to determine our outer and inner radius. So the outer In this case will be given by our topmost function. Inner will be given by our bottommost function. In this case, the top function is the cube root of X, and the bottom function is just X. Now we have our two equations. Let's go ahead and find our balance. We can find our balance by setting the equations equal to each other. We have the cube root of X equals X, or X to the 1/3 power equals x. Let's go ahead and solve. We can raise everything to the 3rd power to get rid of the fractional exponent, giving us X equals X to the 3rd. Now, if we subtract X on both sides, we get X the 3 minus X equals 0. And if we factor out X, we get X multiplied by X2 minus 1 equals 0. We get x equals 0. X2 minus 1 equals 0, which gives us X equals positive and -1. Because we are in the first quadrant. Our A or our lower bounds will be 0, and our upper bounds will just be the positive one. Let's go ahead and write our integral now. We have the integral from 0 to 1 of pi multiplied by X to the 1/3. Squared minus x 2 In this case, we'll have DX because we are in terms of X. Now we can solve our integral. We can simplify once more by moving the pie outside. And multiplying our exponent to get X to the 2/3 minus X2. Now from here, we can take our antiderivatives. We get pi, multiplied. By the inside derivative of X to 2/3, which is 3/5 X 5/3. Minus x the 3 divided by 3. From 0 to 1 Now, plugging in 0 will just give us 0, so we can just plug in 1. Pi multiplied by 3/5 minus 1/3. We can simplify this, giving us 4 pi divided by 15. This is in cubic units, and this makes the answer to our problem answer B. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.y=x and y=4√x; about the x-axis

Watch next

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.
y=x and y=4√x; about the x-axis