Thread: Find the surface area generated by revolving the curve

Need help on finding the surface area generated by revolving the curve

y=square root of 25-x^2 around the X-axis for -3 ≤ x ≤ 4

Thank you

Originally Posted by macadamianchica

Need help on finding the surface area generated by revolving the curve

y=square root of 25-x^2 around the X-axis for -3 ≤ x ≤ 4

Thank you

find it using the definite integral of y in [-3;4]

Originally Posted by macadamianchica

Need help on finding the surface area generated by revolving the curve

y=square root of 25-x^2 around the X-axis for -3 ≤ x ≤ 4

Thank you

As the curve revolves around the x-axis, each point on graph of the function traces out a circle, with f(x) as radius.

To calculate the surface area, the circle circumferences are integrated.



Hence, the curved surface area is....

You may or may not want to include the circular side faces.

Originally Posted by macadamianchica

Need help on finding the surface area generated by revolving the curve

y=square root of 25-x^2 around the X-axis for -3 ≤ x ≤ 4

Thank you

the funct y=sqrt25-x^ is a halfcircle. find the integral by replacing x=5sint t. then u have sqrt of 25-25sin^2 x=sqrt25(1-sin^2 x)=5cost now find the definite integral of y=5cost

This is not the "integration of a function problem".
It's a 3-dimensional "surface of revolution".