Volume of a sphere Let R be the region bounded by the upper ha...

Question

Volume of a sphere Let R be the region bounded by the upper half of the circle x²+y² = r² and the x-axis. A sphere of radius r is obtained by revolving R about the x-axis.

a. Use the shell method to verify that the volume of a sphere of radius r is 4/3 πr³.

Accepted Answer

Welcome back, everyone. In this problem, let Q be the region bounded by the right half of the circle, X2 + Y2 equals R2 and the Y axis. Find the volume of the solid formed when Q is revolved above the y-axis using the shell method. Now first, let's try to visualize our problem to help us solve. So let's draw our X and Y axis, OK. And for our axis, we know that we're looking at a circle, the right half of the circle, X2 + Y2 equals R2, and this equation represents a circle with a radius R that's centered at the origin of our graph. So our circle will probably look something like this, OK? And since it's bounded by the Y axis, then the region Q would be this entire region shaded in red. This is what we're going to revolve about the Y axis. OK. So, let's label our region Q. Now how can we figure out what the volume is when Q is revolved about the y axis from the shell method? What do we know about the shell method? Well, to use the shell method with revolution about the y axis, that is vertically, then we take a vertical slice at a typical X with X greater than or equal to 0 but less than or equal to R. So in other words, at a typical X. OK. Then our small slice or small vertical slice, OK, is going to be equal to the circumference. OK, of our circle. Multiplied by the height of our circle, multiplied by its thickness d X. Now how can we find the circumference and and the height of our circle? Well, given that the radius here. Sorry, given the radius here since we're. Revolving it around the y axis, the radius would be equal to X, OK? Then its height, well, first it's circumference. We know that the circumference of a circle is 2xR, so the circumference in this case will be 2 by X because the radius is X. On the other hand, its height is going to be determined by Y values satisfying Y2 equals R2 minus X2. OK, so the height is going to be from the lower half negative values of Y, OK, which if we solve for Y, Y would be equal to the square root of R2 minus X2. So it would comprise of the lower side of our circle, which would be negative values of Y negative r quad minus X2, and the upper part of our circle, which would be positive values of R2 minus X, the square root of r quad minus X2. I know when we take that entire height together when we combine both of those heights, then from our diagram we can see. That the height, OK, the height is going to be equal to 2 multiplied by a root R2 minus X2. So since we have our circumference. And we have our height then that means our vertical slice the volume of our vertical slice is going to be equal to 2 X multiplied by 2 root R2 minus X2 multiplied by DX which when we simplify would be 4 X root R2 minus X2 D X. Know that we have our vertical slice, then we can integrate this to get our total volume. So integrating. From X equals 0, OK, so X equals R. Our upper limit because it would be our radius then our volume will be equal to the definite integral from 0 tor of 4 pi X multiplied by a root r quad minus X2d with respect to X. And that would be equal to 4 multiplied by the definite integral of x root R2 minus X2 with respect to X. Now we can integrate by using substitution. Now through substitution, let's let you be equal to root R2 minus X2. OK, sorry, it would simply be equal to R2 minus X2. So let U equals R2 minus X2. Then that means the derivative of U with respect to X would be equal to -2 X, which means that we could rewrite XD X. OK, as negative half of the EU. And that helps because notice that we have X and DX in our original term. So now we've been able to, we, we should be able to express this purely in terms of you. The only thing we have left is to try and rewrite our limits in terms of you. Now, when X equals 0 or lower limit using the formula for you, that means you would be equal to R squared. On the other hand, when X equals R, the upper limit, then U equals R2 minus R2, which is 0. So if we substitute this into our original term. This now means in terms of you or volume is going to be equal to 4 multiplied by negative or half or rather let me write that together so you can see it clearly 4 pi multiplied by the root between 0 and R2 OK of. You to the power of a half, that's you or root you that we would have here multiplied by XDXR negative a half the U. And know if we simplify that even further we can bring our constant oath and this is going to give us negative 2 pi multiplied by the definite integral between 0 and R squared of you to the power of a half with respect to you. Now we can integrate. When we integrate here, we're going to have negative 2 pi multiplied by 2/3 of Upo of three halves between 0 and R2d and evaluating evaluating from U equals R2 to U equals 0. for U equals R2, this will be negative 2 multiplied by 2/3, OK, of. 0, well. 2/3 of 0, OK. Well, no, let's factor with about 2/3 here. 2/3 of 0 to the pore of 3 halves minus R squared to the pore of 3 halves. I know if we simplify this even further, this will be -2 pi multiplied by 2/3 of our cubed, OK. And That is equal to 4/3 of pi are cubed. Thus, the volume is 4/3 of prior cubed cubic units. Thanks a lot for watching everyone. I hope this video helped.

Volume of a sphere Let R be the region bounded by the upper half of the circle x²+y² = r² and the x-axis. A sphere of radius r is obtained by revolving R about the x-axis.a. Use the shell method to verify that the volume of a sphere of radius r is 4/3 πr³.

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Volume of a sphere Let R be the region bounded by the upper half of the circle x²+y² = r² and the x-axis. A sphere of radius r is obtained by revolving R about the x-axis.

a. Use the shell method to verify that the volume of a sphere of radius r is 4/3 πr³.