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Christopher Croke University of Pennsylvania Multiple Integrals Math 115 Calculus 115 Integrals. The geometry of integrals of functions of one variable Calculus 115 Rb a f (x)dx: Integrals. The geometry of integrals of functions of one variable For two variables RR R Rb a f (x)dx: f (x, y )dA: Volume under x, y plane counts negative. Calculus 115 How to calculate Use iterated integrals. Calculus 115 How to calculate Use iterated integrals. Z b hZ a h(x) i f (x, y )dy dx g (x) where a and b are constants and g (x) and h(x) are functions of x only. Calculus 115 How to calculate Use iterated integrals. Z b hZ a h(x) i f (x, y )dy dx g (x) where a and b are constants and g (x) and h(x) are functions of x only. How to compute? Calculus 115 How to calculate Use iterated integrals. Z b hZ a h(x) i f (x, y )dy dx g (x) where a and b are constants and g (x) and h(x) are functions of x only. How to compute? First find an antiderivative F (x, y ) thinking of x as a constant, that is: ∂F (x, y ) = f (x, y ). ∂y Calculus 115 How to calculate Use iterated integrals. Z b hZ a h(x) i f (x, y )dy dx g (x) where a and b are constants and g (x) and h(x) are functions of x only. How to compute? First find an antiderivative F (x, y ) thinking of x as a constant, that is: ∂F (x, y ) = f (x, y ). ∂y Then compute: Z b h i F (x, h(x)) − F (x, g (x)) dx. a (which is just an integral of a function of one variable.) Calculus 115 Problem: Compute: Z 0 2hZ 4 i (x 2 y )dy dx. 3 Calculus 115 Problem: Compute: Z 2hZ 4 0 i (x 2 y )dy dx. 3 (You don’t need the [, ]’s) Calculus 115 Problem: Compute: 2hZ 4 Z 0 i (x 2 y )dy dx. 3 (You don’t need the [, ]’s) Problem: Compute: Z 1 Z 1+x xydydx. −1 x2 Calculus 115 How does Z a relate to b hZ g2 (x) i f (x, y )dy dx g1 (x) Z Z f (x, y )dA? R Calculus 115 How does Z a relate to b hZ g2 (x) i f (x, y )dy dx g1 (x) Z Z f (x, y )dA? R It is the answer when R is the region: Calculus 115 Problem: Calculate the volume of the solid bounded above by f (x, y ) = x 2 over the rectangle 0 ≤ x ≤ 1, 2 ≤ y ≤ 3. Calculus 115 Problem: Calculate the volume of the solid bounded above by f (x, y ) = x 2 over the rectangle 0 ≤ x ≤ 1, 2 ≤ y ≤ 3. Problem: Calculate Z Z (x + y )dydx R where R is the region in the triangle with vertices (0, 0), (2, 0) and (0, 1). Calculus 115 Problem: Calculate the volume of the solid bounded above by f (x, y ) = x 2 over the rectangle 0 ≤ x ≤ 1, 2 ≤ y ≤ 3. Problem: Calculate Z Z (x + y )dydx R where R is the region in the triangle with vertices (0, 0), (2, 0) and (0, 1). Sometimes need to split the region into two (or more) pieces. Calculus 115 Problem: Calculate the volume of the solid bounded above by f (x, y ) = x 2 over the rectangle 0 ≤ x ≤ 1, 2 ≤ y ≤ 3. Problem: Calculate Z Z (x + y )dydx R where R is the region in the triangle with vertices (0, 0), (2, 0) and (0, 1). Sometimes need to splitR the R region into two (or more) pieces. Problem *: Calculate R y dA where R is the region in the triangle with vertices (−1, 0),(1, 0), and (0, 1). Calculus 115 Often better to integrate w.r.t. x first. Fubini’s Theorem says we get the same answer. Calculus 115 Often better to integrate w.r.t. x first. Fubini’s Theorem says we get the same answer. Z d hZ c h2 (y ) i f (x, y )dx dy h1 (y ) corresponds to the region Calculus 115 Easy example: Z 1Z 2 Z 2Z 1 (x + y )dydx = 0 0 (x + y )dxdy . 0 0 Calculus 115 Easy example: Z 1Z 2 Z 2Z 1 (x + y )dydx = 0 0 (x + y )dxdy . 0 0 Now we will do Problem* this way. Calculus 115 Easy example: Z 1Z 2 Z 2Z 1 (x + y )dydx = 0 0 (x + y )dxdy . 0 0 Now we will do Problem* this way. Sometimes you *have* to do it this way!: Problem: Compute: Z 1Z 1 2 e y dydx. 0 x Calculus 115 Easy example: Z 1Z 2 Z 2Z 1 (x + y )dydx = 0 0 (x + y )dxdy . 0 0 Now we will do Problem* this way. Sometimes you *have* to do it this way!: Problem: Compute: Z 1Z 1 2 e y dydx. 0 x There is no good rule to tell you which way to go. Often the shape of the region gives a hint. Calculus 115 Problem: Find the volume of the solid in the first octant bounded by the coordinate planes, the cylinder x 2 + y 2 = 1, and the plane z − 2y = 0. Calculus 115 Problem: Find the volume of the solid in the first octant bounded by the coordinate planes, the cylinder x 2 + y 2 = 1, and the plane z − 2y = 0. Sometimes can integrate over unbounded regions: Calculus 115 Problem: Find the volume of the solid in the first octant bounded by the coordinate planes, the cylinder x 2 + y 2 = 1, and the plane z − 2y = 0. Sometimes can integrate over unbounded regions: Problem: Compute: Z 1 ∞Z 1 x y 3 xdydx. 0 Calculus 115 Problem: Find the volume of the solid in the first octant bounded by the coordinate planes, the cylinder x 2 + y 2 = 1, and the plane z − 2y = 0. Sometimes can integrate over unbounded regions: Problem: Compute: Z 1 ∞Z 1 x y 3 xdydx. 0 Calculus 115 Riemann sums Break your rectangular region into smaller rectangles Rk of area A(Rk ). Calculus 115 Riemann sums Break your rectangular region into smaller rectangles Rk of area A(Rk ). Choose a point (xk , yk ) in each Rk . Calculus 115 Riemann sums Break your rectangular region into smaller rectangles Rk of area A(Rk ). Choose a point (xk , yk ) in each R R Rk . Then the Riemann sum that approximates the integral R f (x, y )dA is: Σk f (xk , yk )A(Rk ). Calculus 115 Riemann sums Break your rectangular region into smaller rectangles Rk of area A(Rk ). Choose a point (xk , yk ) in each R R Rk . Then the Riemann sum that approximates the integral R f (x, y )dA is: Σk f (xk , yk )A(Rk ). Calculus 115 Riemann sums Break your rectangular region into smaller rectangles Rk of area A(Rk ). Choose a point (xk , yk ) in each R R Rk . Then the Riemann sum that approximates the integral R f (x, y )dA is: Σk f (xk , yk )A(Rk ). Calculus 115 There is a theorem which says that the limit as the size of the rectangles Rk goes to 0 then the Riemann sum approaches the integral. Calculus 115 There is a theorem which says that the limit as the size of the rectangles Rk goes to 0 then the Riemann sum approaches the integral. You can see from the picture why Fubini’s theorem works. Calculus 115 There is a theorem which says that the limit as the size of the rectangles Rk goes to 0 then the Riemann sum approaches the integral. You can see from the picture why Fubini’s theorem works. You can do the same sort of thing if the region is not a rectangle. Calculus 115 There is a theorem which says that the limit as the size of the rectangles Rk goes to 0 then the Riemann sum approaches the integral. You can see from the picture why Fubini’s theorem works. You can do the same sort of thing if the region is not a rectangle. END OF MATERIAL FOR THE MIDTERM Calculus 115