{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lec_week7

lec_week7 - MIT OpenCourseWare http/ocw.mit.edu 18.02...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . 18.02 Lecture 16. – Thu, Oct 18, 2007 Handouts: PS6 solutions, PS7. Double integrals. Recall integral in 1-variable calculus: a b f ( x ) dx = area below graph y = f ( x ) over [ a, b ]. Now: double integral R f ( x, y ) dA = volume below graph z = f ( x, y ) over plane region R . Cut R into small pieces Δ A the volume is approximately f ( x i , y i ) Δ A i . Limit as Δ A ⇒ → gives R f ( x, y ) dA . (picture shown) How to compute the integral? By taking slices: S ( x ) = area of the slice by a plane parallel to yz-plane (picture shown): then x max volume = S ( x ) dx, and for given x , S ( x ) = f ( x, y ) dy. x min In the inner integral, x is a fixed parameter, y is the integration variable. We get an iterated integral. Example 1: z = 1 − x 2 − y 2 , region ≤ x ≤ 1 , ≤ y ≤ 1 (picture shown): 1 1 (1 − x 2 − y 2 ) dy dx....
View Full Document

{[ snackBarMessage ]}

Page1 / 4

lec_week7 - MIT OpenCourseWare http/ocw.mit.edu 18.02...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online