In this exercise we have to use the knowledge of volume of a solid and through integration calculate this volume, then:
[tex]V=\int\limits^5_{-5} {4({25-x})} \, dx \\[/tex]
In this way we will use the equation of the circle of radius 5 and then write this equation as a function of Y, this will be:
[tex]x^2+y^2=5^2\\y^2=25-x^2\\y=+\sqrt{25-x} \\[/tex]
Now to find the volume we will integrate as a function of Y, this will be:
[tex]V=\int\limits^5_{-5} {(2\sqrt{25-x})^2} \, dx \\V=\int\limits^5_{-5} {(4({25-x})} \, dx \\[/tex]
See more about volume at brainly.com/question/1578538
