Question

Calculate the volume of the solid generated by spining the region delimited by the next functions around the x-axis:?

`y^2-x^2=1, y=2`

Answer 1

`3.464pi` or 10.883

Explanation:

Well since it will be rotating around the x-axis you want to isolate y in your equation. So `y^2-x^2=1` can be changed to `y=sqrt(1+x^2)`.

Next you will want to graph the equation to be able to evaluate if you will be using the washer or dish method(basically to see if there are holes in the solid).
graph{sqrt(1+x^2) [-2.792, 7.21, -1.12, 3.88]}

Now we know that we will be using the washer method. Next, we need to find the intersection point of our equation and y=2. Using a graphing calculator, or by plugging 2 into y, we know that the intersection point will be at 1.732. Now we can start the integral function.

`piint_0^(1.732)(2)^2-(sqrt(1+x^2))^2dx`
2 is the large radius of the shape and the function is the area under the curve that you need to subtract. And remember that the base volume for this is `V=pir^2`.

To calculate this we first should simplify to equation. It will become `piint_0^(1.732)(3-x^2)dx`

Now we need to find the antiderivative of the inside function. So the derivative of `3-x^2` is `3x-x^3/3`

All you need to do now is apply the A and B functions to that equation(A being 0 and B being 1.723)(you plug b into x and subtract that from a plugged into x).
You'll get 3.464
Don't forget to multiply by `pi`!
Your answer is now `3.464pi` or 10.883.