$y=(25-x^2)^{1/2},\;y=3$ find the volume

Question

Here is the question if $y=(25-x^2)^{1/2},$ $y=3$ find the volume?(about x-axis)

I draw this and I have circle and line But the problem is when y=3 we will have two area and both of them are enclosed by the given curve Wich of them should I choose ? Thanks all

Answer 1

Be careful that the below part is not closed by them, and only above part of the line is closed by them. $$ V=\pi\int_{-4}^4\left(\left(\sqrt{25-x^2}\right)^2-3^2 \right)\,dx $$

Answer 2

The desire integral is indeed as follows: $$\pi\int_{-4}^{+4}(y_1^2-y_2^2)dx$$ wherein $y_1$ is the upper curve along side the interval $I [-4,4]$ and $y_2$ is the downer one. All we are doing here is based on this method. The following plot shows the volume of revolution: