The region R is bounded by the graph of f(x)=2x(2−x) and the x...

Question

The region R is bounded by the graph of f(x)=2x(2−x) and the x-axis. Which is greater, the volume of the solid generated when R is revolved about the line y=2 or the volume of the solid generated when R is revolved about the line y=0? Use integration to justify your answer.

Accepted Answer

Let regions be bounded by the graph of G of X equals 3 X multiplied by 3 minus X and the X axis, which is greater the volume of the solid generated when S is revolved about the line Y equals 27 divided by 4, or when S is evolved about the line Y equals 0. Now to solve this, we'll have to find two integrals. Let's first find the bounds of these integrals. What we can do is find the X intercepts of our equation G of X. We can say 3 X multiplied by 3 minus X equals 0. Giving us 3 X equals 0 and 3 minus X equals 0. Gimmicks values of X equals 0 and X equals 3. So now we know our equation. Is going to be bounded between 0 and 3. Let's set up our intervals. We'll first use the disk method. It will do this for the revolution around Y equals 0. We have the integral From A to B, a pi radius square. In this case, this will be DX. We can then write this as the integral. From 0 to 3 of Pi, multiplied by 3 X, multiplied by 3 minus X D X. Now we just need to solve. And this is all squared. We can expand the term inside with the squared, we end up getting pi multiplied by the integral. Of 81 X2. 54 execute. Plus 9 x to the 4th. Now we just take our integral. We get paid, multiplied. By 81 X the 3rd divided by 3. Minus 54 X of the fourth divided by 4. Plus 9 x 5th divided by 5. From 0 to 3. Now, plugging in 0 is just going to give us 0 everywhere. So, we can just plug in 3 for the center go. When we plug in 3, we end up getting the final value. A pi multiplied by 729 divided by 10. We can just write this as 729 divided by 10 pi. Cubic units. Now, the problem's goal is to determine which one gives us a larger volume, so we'll also check another integral. So, we have the integral now. As you go from 0 to 3, revolve around the line Y equals 27/4. Will rewrite this integral. As the integral from A to B. Of pi multiplied. Bye. The outside radius squared minus the inside radius squared. The exigant Now, we have the integral from 0 to 3 of pi, multiplied by our outside radius, which ends up being 27 divided by 4 squared. Minus our inside radius, 27 divided by 4, minus 3X multiplied by 3 minus X squared. Let's go ahead and expand and rewrite this. When expanded and rewritten, We ended up getting the integral of pi, multiplied. Bye 243, divided by 2. X Minus 243 divided by 2 X2. Plus 54. Executes minus 9 X to the 4th. To find that, you just expand and simplify our algebra. We can take the antiderivative. We will get pie multiplied. By 243, divided by 2. X 2 divided by 2 minus 243 halves, multiplied by XQ divided by 3, plus 54 multiplied by X of the fourth divided by 4, minus 9 X to the 5th divided by 5, from 0 to 3 once again. Plugging in 0 will once again give us 0, so we can just plug in 3. This results in the value of pi, multiplied. By 2187 divided by 20. Which is just 2187 divided by 20 pi cubic units. We then see Our volume of 2187 divided by 20 pi is larger than our other volume of 729 divided by 10 pi. We can then say The volume Of S Revolved Around Y equals 27 divided by 4. Is greater This is the answer to our problem. OK, I hope that help you solve the problem. Thank you for watching. Goodbye.

The region R is bounded by the graph of f(x)=2x(2−x) and the x-axis. Which is greater, the volume of the solid generated when R is revolved about the line y=2 or the volume of the solid generated when R is revolved about the line y=0? Use integration to justify your answer.

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