A torus (doughnut) A torus is formed when a circle of radius 2...

Question

A torus (doughnut) A torus is formed when a circle of radius 2 centered at (3, 0) is revolved about the y-axis.

b. Use the washer method to write an integral for the volume of the torus.

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says a torus is generated by revolving a circle of radius 3 centered about 4.0 about the y axis, right in an integral that represents the volume of the torus using the washer method. So the first thing we want to do is create a sketch of our circle. And so we're going to draw our X and Y axis so that we focus on the 1st and 4th quadrant. And we're told that our circle is centered at 4.0, so we'll put a center point there, and we're told that it has a radius of 3. So, we'll sketch a circle that has a radius of 3 centered about 4.0. And we're going to revolve the area that is enclosed by that circle about the Y axis to form a torus. So, if we look at our sketch here, we see that our circle is symmetric about the X-axis. And since we're revolving about the y-axis using the washer method, we're going to want to integrate with respect to Y. So that means that we need to write X as a function of Y. So we're going to begin using our equation for a circle. So we call that for a circle centered at 4.0 and a radius 3. That equation is given by the quantity of X minus 4 in quantity squared, plus Y2 is equal to 32, and we need to solve this equation for X. So to begin with, we'll subtract Y2d from both sides of the equation. So we'll have the quantity of X minus 4 in quantity squared is equal to 9 minus Y2. Taking the square root of both sides, we'll have that X minus 4 is equal to plus or minus the square root of the quantity of 9 minus Y2 in quantity. Adding 4 to both sides, we'll get that X is equal to 4 + or minus the square root of the quantity of 9 minus y2 in quantity. So, based upon this information, we see that our circle is bounded on the right by X is equal to 4, plus the square root of quantity of 9 minus Y2d, and it's bounded on the left by X equals to 4 minus the square root of quantity of 9 minus Y2. So now we have enough information to create our integral for our volume. So we call that for the washer method, our volume V is going to be equal to i multiplied by the interval from A to B, of the quantity of our outer radius squared minus our inner radius squared in quantity D Y. So, here for our outer radius, that is just going to be the X curve that bounds our region from the right, and the inner radius is going to be the curve that bounds our region from the left. So that means that our volume V is going to be equal to pi, multiplied by the interval from -3 to 3. Since our circle spans the Y values going from -3 to 3. Of the quantity of the quantity and herefore are out we'll have 4 plus the square root of the quantity of 9 minus y2 in quantity and that whole quantity is squared. Then we'll have minus our inner radius squared, so we'll have the quantity of 4 minus the square root of the quantity of 9 minus Y squared in quantity, and that whole quantity is squared in quantity D Y. So we want to do is multiply everything out inside the integran. In doing so, we see our volume V is going to be equal to pi, multiplied by the interval from -3 to 3. Of the quantity of 16 + 8 multiplied by the square root of the quantity of 9 minus Y2 in quantity. Plus 9 minus y2d. -16. Plus 8 multiplied by the square root of the quantity of 9 minus y2 in quantity minus 9 plus y2 in quantity D Y. And when we simplify this this expression, we see that this is going to be equal to pi multiplied by the interval from -3 to 3 of 16 multiplied by the square root of the quantity of 9 minus y squad in quantity D Y. Now 16 is a constant, so we can pull it out front of the interval. In doing so, we see that our volume V is going to be equal to 16 pi multiplied by the integral from -3 to 3 of the square root of the quantity of 9 minus y2d and quantity D Y, and 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.

A torus (doughnut) A torus is formed when a circle of radius 2 centered at (3, 0) is revolved about the y-axis.

b. Use the washer method to write an integral for the volume of the torus.

Key Concepts

Washer Method for Volume

Setting up the Integral with Respect to y-axis

Equation of the Circle and Its Position

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