45.4k views
2 votes
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
√(121 - x^2) to the semicircle y equals
√(121 - x^2). Find the volume of the solid.

User Snorkpete
by
7.7k points

1 Answer

3 votes

Answer:

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

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.

A solid lies between planes perpendicular to the​ x-axis at x = -11 and x = 11. The-example-1
User Luke Duda
by
8.0k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories