Question

A solid lies between planes perpendicular to the​ x-axis at x = -11 and x = 11. The​ cross-sections perpendicular to the​ x-axis between these planes are squares whose bases run from the semicircle y equals negative $\sqrt{121 - x^2}$ to the semicircle y equals $\sqrt{121 - x^2}$. Find the volume of the solid.

Answer 1

Answer:

The answer for the volume of the solid is 7098.67 unit^3.

Step-by-step explanation:

As mentioned in the question semicircle y Equals

y=−√121−x^2

to the semicircle

y=√121−x^2

Base of square is,

B=2√121−x^2

Area of square:

A=b^2

Substitute:

A=(2√121−x^2)^2

=4(121−x^2)

limits are from:

−11 to 11.

Expression since the limits are -11 and 11.