DOUBLE INTEGRALS MATH 200 GOALS Be able to compute double integral - PowerPoint PPT Presentation

MATH 200 WEEK 7 - FRIDAY DOUBLE INTEGRALS

MATH 200 GOALS ▸ Be able to compute double integral calculations over rectangular regions using partial integration. ▸ Know how to inspect an integral to decide if the order of integration is easier one way (y first, x second) or the other (x first, y second). ▸ Know how to use a double integral as the volume under a surface or find the area or a region in the xy-plane. ▸ Be able to compute double integral calculations over non rectangular regions using partial integration.

MATH 200 INTEGRATION IN CALC II � b n � f ( x ∗ lim k ) ∆ x = f ( x ) dx = F ( b ) − F ( a ) n →∞ a k =1 RIEMANN SUM FUNDAMENTAL THEOREM OF CALCULUS

MATH 200 DOUBLE INTEGRALS ▸ Rather than integrating over Surface: f(x,y) intervals on the x-axis (or y- axis), we will be integrating over regions in the plane . ▸ Given a region R, divide it up into squares with area Δ A = Δ x Δ y (x k ,y k ) ▸ Pick a point (x k ,y k ) in each Region: R square Area of each is Δ x Δ y ▸ Add the products f(x k ,y k ) Δ x Δ y EACH PRODUCT IS THE VOLUME OF A RECTANGULAR PRISM

MATH 200 ▸ Skipping ahead to definite integrals… Surface: f(x,y) ▸ The definite integral of f(x,y) over the region R Region: R yields the net-signed volume bounded between R and the surface z=f(x,y)

MATH 200 AN EXAMPLE Surface: f(x,y) ▸ Consider the paraboloid f(x,y) = x 2 + y 2 over the rectangular region R = {(x,y) : 0 ≤ x ≤ 1 & 1 ≤ y ≤ 3} ▸ One way to set up a double (iterated) integral for the net-signed volume bounded between R and the surface is like this: Region: R � 3 � 1 ( x 2 + y 2 ) dx dy This is the (net-signed) 1 0 volume we’re finding

MATH 200 OKAY, NOW HOW DO WE ACTUALLY INTEGRATE THAT ▸ This integral can be viewed as one integral nested in another: � 3 � 1 � 3 �� 1 � ( x 2 + y 2 ) dx dy = ( x 2 + y 2 ) dx dy ⇒ 1 0 1 0 ▸ For the inner integral, x is our variable, so we’ll take the partial antiderivative with respect to x and evaluate at x=0 and x=1: � 3 �� 1 � 3 � � 1 x 3 � � ( x 2 + y 2 ) dx 3 + xy 2 � dy = dy � � 1 0 1 0 � 3 � 1 � 3 + y 2 − (0 + 0) = dy 1 � 3 � 1 � 3 + y 2 = dy 1

MATH 200 ▸ Now we just have a Calc II integral to evaluate: � 3 3 3 y + y 3 � � 1 � dy = 1 � 3 + y 2 � 3 � 1 1 � 1 3 + 1 � = (1 + 9) − 3 = 28 3 ▸ So that’s the net-signed volume bounded between f(x,y) and R ▸ We could have set the integral up the opposite way as well… � 1 � 3 ( x 2 + y 2 ) dy dx 0 1

MATH 200 � 1 � 3 � 1 � � 3 x 2 y + y 3 � ( x 2 + y 2 ) dy dx = � dx � 3 � 0 1 0 1 � 1 � � �� x 2 + 1 3 x 2 + 9 − = dx 3 0 � 1 � � 2 x 2 + 26 = dx 3 0 1 = 2 x 3 � + 26 � 3 x � 3 � 0 � 2 � 3 + 26 = − (0 + 0) 3 = 28 3

MATH 200 LET’S TRY ANOTHER ONE � 2 � π � π 2 x 2 � � x sin y dx dy = 2 sin y dy � � 0 1 0 1 � π 2 sin y − 1 = 2 sin y dy 0 � π = 3 sin y dy 2 0 π � = − 3 � 2 cos y � � 0 = − 3 2( − 1 − 1) = 3

MATH 200 NON RECTANGULAR REGIONS ▸ We’ll divide non rectangular regions into two types: ▸ TYPE 1: Regions bounded below and above by curves y 1 (x) and y 2 (x) from x=a and x=b ▸ TYPE 2: Regions bounded on the right and left by two curves x 1 (y) and x 2 (y) from y=c and y=d

MATH 200 TYPE 1 REGIONS ▸ Order of integration: dydx � b � y 2 ( x ) �� f ( x, y ) dA = f ( x, y ) dydx y 1 ( x ) a R x=a x=b y 2 (x) y 1 (x)

MATH 200 TYPE 2 REGIONS ▸ Order of integration: dxdy � d � x 2 ( y ) �� f ( x, y ) dA = f ( x, y ) dxdy x 1 ( y ) c R y=d x 1 (y) x 2 (y) y=c

MATH 200 EXAMPLE ▸ Say we want to find the volume bounded between f(x,y)=xy 2 and the xy-plane over the region bounded by y=x and y=x 2

MATH 200 ▸ Let’s look at the region on the xy-plane: ▸ We can treat this as a Type 1 region: ▸ It’s bounded below and above by y 1 (x) = x 2 and y 2 (x) = x ▸ The region extends from x = 0 to x = 1 � 1 � x �� x 2 xy 2 dydx f ( x, y ) dA = 0 R

MATH 200 � 1 � 1 � � � x y = x � 1 TAKE ANTIDERIVATIVE x 2 xy 2 dydx = 3 xy 3 � dx � WITH RESPECT TO Y � 0 0 y = x 2 � 1 � 1 ( x ) 3 − ( x 2 ) 3 �� PLUG IN Y=X AND � = 3 x dx Y=X 2 AND SUBTRACT 0 � 1 � 1 x 3 − x 6 �� SIMPLIFY � = 3 x dx 0 � 1 = 1 x 4 − x 7 � � dx 3 0 1 �� � 1 = 1 5 x 5 − 1 INTEGRATE WITH 8 x 8 � � RESPECT TO X 3 � 0 � 3 � = 1 3 40 = 1 40

MATH 200 ▸ We also could have treated this as a Type 2 region ▸ It’s bounded to the right and to the left x 1 (y)=y and x 2 (y)=y 1/2 ▸ The region extends from y=0 to y=1 � 1 � y 1 / 2 �� xy 2 dxdy f ( x, y ) dA = y 0 R

MATH 200 � y 1 / 2 x = y 1 / 2 � 1 � 1 � � � TAKE ANTIDERIVATIVE 1 xy 2 dydx = 2 y 2 x 2 � dx WITH RESPECT TO X � � y 0 0 x = y PLUG IN X- x = y 1 / 2 � 1 � � ( y 1 / 2 ) 2 − ( y ) 2 �� 1 2 y 2 � BOUNDS AND � = dx � SUBTRACT � 0 x = y � 1 � 1 y − y 2 �� 2 y 2 � = dx SIMPLIFY 0 � 1 = 1 y 3 − y 4 � � dx 2 0 1 �� � 1 = 1 4 y 4 − 1 INTEGRATE WITH 5 y 5 � � 2 RESPECT TO Y � 0 � 1 � = 1 2 20 = 1 40

MATH 200 ANOTHER EXAMPLE ▸ Let’s integrate the same function over a different region. ▸ Let f(x,y)=xy 2 and take R to be the region bounded by y=3x, y=x/2 and y=1 THE VOLUME WE’RE FINDING REGION OVER WHEN WE WHICH WE’RE INTEGRATE INTEGRATING

MATH 200 ▸ We want to set the limits of integration to cover the region R ▸ Notice that we can’t set this up as a single Type 1 region ▸ It’s two Type 1 regions: ▸ (1) y 1 (x)=x/2; y 2 (x)=3x ▸ From x=0 to x=1/3 ▸ (2) y 1 (x)=x/2; y 2 (x)=1 ▸ From x=1/3 to x=2

MATH 200 ▸ Setup as Type 1 region: � 1 / 3 � 3 x � 2 � 1 �� xy 2 dydx + xy 2 dydx f ( x, y ) dA = x/ 2 1 / 3 x/ 2 0 R ▸ How about setting this up as a Type 2 region?

MATH 200 ▸ The region is bounded on the left by x 1 (y) = y/3 ▸ Solve y=3x for x ▸ The region is bounded on the right by x 2 (y) = 2y ▸ Solve y=x/2 for x ▸ The region extends from y=0 to y=1 � 1 � 2 y �� xy 2 dxdy f ( x, y ) dA = y/ 3 0 R

MATH 200 � 1 � 2 y � 1 2 y � 1 xy 2 dxdy = 2 y 2 x 2 � dy � � y/ 3 0 0 y/ 3 � 1 � � 2 � 1 � y 2 y 2 (2 y ) 2 − = dy 3 0 � 1 = 1 35 9 y 4 dy 2 0 1 �� � 1 = 35 5 y 5 � � 18 � 0 = 7 18(1 − 0) = 7 18

MATH 200 A TRICKIER EXAMPLE ▸ Consider the double integral � √ π / 2 � √ π sin( x 2 ) dxdy 2 y 0 ▸ We can’t integrate sin(x 2 ) with respect to x! ▸ But, we can try to switch the order of integration to dydx ▸ Let’s look at what’s going on with the region R: ▸ For the x-bounds we have x 1 (y) = 2y and x 2 (y) = sqrt( π ) ▸ x = 2y has the same graph as y = x/2 ▸ For the y-bounds we have y 1 = 0 and y 2 = sqrt( π )/2

MATH 200 ▸ So R is bounded on the left by x 1 (y) = 2y and on the right by x 2 (y) = sqrt( π ) from y=0 to y=sqrt( π )/2 ▸ How can we set this up as a Type 2 region? ▸ R is bounded on the bottom by y 1 (x) = 0 and on top by y 2 (x) = x/2 ▸ In the x-direction, it extends from x 1 = 0 to x 2 = sqrt( π )

MATH 200 ▸ Now the question is, “Can we integrate this as a dydx integral?” � x/ 2 y = x/ 2 � √ π � √ π � � sin( x 2 ) dydx = sin( x 2 ) y dydx � � 0 0 0 y =0 � √ π � x � sin( x 2 ) = 2 − 0 dydx 0 � √ π = 1 x sin( x 2 ) dydx 2 0 ▸ We can use u-substitution! u = x 2 ; du = 2 x dx ⇒ u = 0; x = √ π = x = 0 = ⇒ u = π ;

MATH 200 � x/ 2 y = x/ 2 � √ π � √ π � sin( x 2 ) dydx = sin( x 2 ) y � dx � � 0 0 0 y =0 � √ π � x � sin( x 2 ) = 2 − 0 dx 0 � √ π = 1 x sin( x 2 ) dx 2 0 � π = 1 2 · 1 sin( u ) du 2 0 � π = − 1 � 4 cos( u ) � � 0 = − 1 4( − 1 − 1) = 1 2

MATH 200 WHAT HAPPENS WHEN WE INTEGRATE 1 ▸ Let f(x,y) = 1 and let R be the region bounded by y=4-x 2 , y=3x, and x=0 in the first quadrant.

MATH 200 ▸ Setting this up as a Type 1 integral, we get � 1 � 4 − x 2 � 1 y =4 − x 2 � � 1 dydx = y dx � � 0 3 x 0 y =3 x � 1 (4 − x 2 ) − (3 x ) � � = dx 0 1 � = 4 x − 1 3 x 3 − 3 2 x 2 � � � 0 = 4 − 1 3 − 3 2 = 13 6

MATH 200 SO…? � 1 ▸ Look at the second line: (4 − x 2 ) − (3 x ) � � dx 0 ▸ From Calculus II we know that this integral yields the net-signed area bounded between y=4-x 2 and y=3x ▸ So the volume of the solid bounded between f and R is 13/6 units 3 BUT it’s also the case that the area bounded between y=4-x 2 and y=3x is 13/6 units 2 ▸ In general… �� 1 dA = Area of R R

MATH 200 ▸ Why does this work? A ▸ For a cylindrical surface with a base of area A and a h V = Ah height of h, the volume is Ah ▸ In the case that h = 1, V = A ▸ NOTE: V=A numerically, but the units are not the same h=1