Question

Find the volume of the solid torus formed when the circle of radius 4 centered at (0,6) is revolved about the \(x\) -axis.

Short answer

Question: Find the volume of the solid torus formed by revolving the circle with radius 4 centered at the point (0,6) around the x-axis. Answer: The volume of the solid torus is \(\frac{800\pi}{3}\) cubic units.

Step-by-step solution

Setting up the Integral

Set up the integral for the volume of the torus. We will be integrating with respect to y, so we need to express the circle equation in terms of y. The circle is centered at (0,6) with radius 4, so its equation is: \((x-0)^2+(y-6)^2=4^2\) Solving for x, we get: \(x=\sqrt{16-(y-6)^2}\)

Determine Inner and Outer Radii

To apply the washer method, we need to determine the inner radius (R) and the outer radius (r) as functions of y. The outer radius, r, is the distance from the x-axis to the far edge of the circle (the point on the circle closest to the x-axis). Since the center of the circle is at (0,6), this point is (0,2), making the outer radius dependent on y as follows: \(r(y)=6-y\) The inner radius, R, is the distance from the x-axis to the point on the circle farthest from the x-axis (the point on the circle closest to the x=0 line). Since the circle is centered at (0,6) and has a radius of 4, the inner radius is the difference between the y-coordinate of the center and the circle's radius. This makes the inner radius constant as follows: \(R=6-4=2\)

Applying the Washer Method

Now we can apply the washer method formula to find the volume. The washer method involves integrating the difference of the outer radius squared and the inner radius squared, multiplied by pi, with respect to the variable of integration (in this case, y): \(V=\pi\int_a^b [r(y)^2-R^2] dy\) We need to determine the limits of integration (a and b), which represent the range of y values covered by the circle. Since the circle is centered at (0, 6) with a radius of 4, it covers y values from 2 to 10: \(a=2, b=10\) Now we can plug in the values we found for r(y), R, a, and b, and evaluate the integral: \(V=\pi\int_2^{10} [(6-y)^2-2^2] dy\)

Evaluating the Integral

To evaluate the integral, first, expand the terms inside to simplify the expression: \(V=\pi\int_2^{10} (36-12y+y^2-4) dy\) Next, integrate each term with respect to y: \(V=\pi \left[36y - 6y^2 + \frac{1}{3}y^3 - 4y \right]_2^{10}\) Now, calculate the result by plugging in the limits of integration: \(V=\pi [(36\cdot 10 - 6\cdot 10^2 + \frac{1}{3}\cdot 10^3 - 4\cdot 10) - (36\cdot 2 - 6\cdot 2^2 + \frac{1}{3}\cdot 2^3 - 4\cdot 2)]\) \(V=\pi[360-304+ \frac{1000}{3} - 40]\) After simplifying, we get: \(V=\frac{800\pi}{3}\) Therefore, the volume of the solid torus is \(\frac{800\pi}{3}\) cubic units.