Iterated Integrals. f (x; y) dy dx. p(x) To evaluate a type I integral, we rst evaluate the inner integral Z q(x) f (x; y) dy.

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1 Iterted Integrls Type I Integrls In this section, we begin the study of integrls over regions in the plne. To do so, however, requires tht we exmine the importnt ide of iterted integrls, in which inde nite integrls re the integrnd of de nite integrl. To begin with, we de ne type I iterted integrl to be n integrl of the form p(x) f (x; y) dy dx To evlute type I integrl, we rst evlute the inner integrl Z q(x) p(x) f (x; y) dy treting x s constnt. We then evlute the result with respect to x: Z " b Z # q(x) f (x; y) dy dx f (x; y) dy dx p(x) p(x) EXAMPLE Evlute the type I integrl Z Z x xy + dydx Solution: To begin with, we integrte with respect to y: Z x xy + dy x y3 x 3 + y x x x3 3 + x As result, we hve Z Z x 3 x4 + x xy + Z dydx 3 x4 + x dx x x + 7

2 Often we evlute the innermost integrl inside the integrnd of the outer integrl rther thn writing the integrtions seprtely. EXAMPLE Evlute the type I integrl Z Z x x ydydx Solution: We rst evlute the inner integrl: Z Z x Z Z x x ydydx x ydy dx Z x y x dx Z x x Z 8 5 x 4 x x dx dx Check your Reding: Why is 5 the denomintor of the result in exmple? Type II Integrls Similrly, we de ne type II integrl to be n iterted integrl of the form Z d Z v(y) c u(y) f (x; y) dxdy It is evluted by considering y to be constnt in the innermost integrl, nd then integrting the result with respect to y. EXAMPLE 3 Evlute the type II integrl Z Z y y (x + y) dxdy

3 Solution: We tret y s constnt in the innermost integrl: Z Z y Z Z y (x y) dxdy (x y) dx dy y y Z hx xy i yy dy Z Z h y 4 y y y 3 dy y i y y dy EXAMPLE 4 Evlute the type II integrl Z Z y sin (y) dxdy Solution: Since we tret y s constnt in the innermost integrl, the function sin (y) cn be considered constnt nd Z Z y Z Z y sin (y) dxdy sin (y) dx dy Z [y sin (y)] dy We now use integrtion by prts with u y nd dv sin (y) dy to obtin u y dv sin (y) dy du dy v cos (y) As result, we hve Z Z [y sin (y)] dy y cos (y)j + cos (y) dy Z Z y sin (y) dxdy cos () Check your Reding: Why cn we write R y sin (y) dx s sin (y) R y dx? Volumes of Solids over Type I Regions 3

4 Let g; h be continuous on [; b] nd supppose tht g (x) h (x) for x in [; b]. If R is region in the xy-plne which is bounded by the curves x ; x b; y g (x), nd y h (x), then R is sid to be type I region. Let s nd the volume of the solid between the grph of f (x; y) nd the xy-plne over type I region R when f (x; y) : To do so, let s notice tht if the solid is sliced with plne prllel to the xz-plne, then its re is A (x) Z h(x) g(x) f (x; y) dy 4

5 It follows tht if fx j ; t j g, j ; : : : ; n, is tgged prtition of [; b] ; then the volume of the solid under the grph of f (x; y) nd over the region R is V nx A (t j ) x j j 5

6 A limit of such simple function pproximtions yields the volumes by slicing formul V Z b A (x) dx 6

7 which is illustrted below: After combining this with the de nition of A (x) ; the result is the iterted integrl Z " b Z # h(x) V f (x; y) dy dx () g(x) EXAMPLE 5 Find the volume of the solid under the grph of f (x; y) x y over the type I region x y x y x Solution: According to (), the volume of the solid is Z Z x V x y dy dx 7

8 We evlute the resulting type I iterted integrl by rst evluting the innermost integrl:d V 3 Z Z y x x y 3 x y 3 dx 4 3 x3 dx Check your Reding: Why is x y non-negtive over the region bounded by x ; x ; y ; y x? Explin. Volumes of Solids over Type II Regions 8

9 Similrly, if p (y) q (y) for y in [c; d], then the region R in the xy-plne bounded by the curves y c; y d; x p (y), nd x q (y), is sid to be type II region. Correspondingly, if f (x; y) for ll (x; y) in type II region R; then the volume of the solid under z f (x; y) nd over the region R is V Z d Z q(y) c p(y) f (x; y) dxdy () EXAMPLE 6 Find the volume of the solid under the grph of f (x; y) x + y over the type II region y x y y x y Solution: To do so, we use () to see tht V Z Z y y x + y dxdy Evluting the innermost integrl leds to Z " # x 3 y V 3 + xy dy y Z 4 3 y3 3 y6 y 4 dy 3 35 Finlly, let us note tht unbounded regions cn led to convergent improper integrls. Indeed, unbounded solids cn hve nite volume. 9

10 EXAMPLE 7 Find the volume of the solid under the grph of f (x; y) e x y over the rst qudrnt. Solution: In the rst qudrnt, x is in (; ) nd y is in (; ) : Thus, () implies tht V Z Z e x y dydx The inner integrl is evluted s n improper integrl V Z Z R lim e x R! Z Z y dydx lim e x e x R dx R! e x dx The resulting integrl is lso evluted s n improper integrl, leding to Z S V lim e x dx lim S! S! e e R Exercises

11 Identify ech integrl s either type I or type II nd evlute: R R R R R 4 R sec(x) tn(x) R R R exp(x) R R (x + y) dydx. R R 3 x y dydx R 3 xy dxdy 4. R R 3 R dydx x x + y R R sin(x) R dydx 6. dydx cos (x) dydx 8. R R x sin (y) dydx R R sin(x) dydx. ydydx R x sin (x) dydx. R R y ex+y dxdy R R y xdydx 4. sin y dxdy R y ln y + R dxdy 6. x ey dxdy R 3 7. R R x x x +y dydx 8. R R x x +y dydx Sketch the region R nd determine its type. Then nd the volume of the solid under z f (x; y) nd over the given region. 9. f (x; y) x + y. f (x; y) 3 R: y ; y R: x ; x x ; x y ; y 4. f (x; y) 3x + y. f (x; y) 6x + y R: x ; x R: x ; x 3 y ; y x y ; y e x 3. f (x; y) xy 4. f (x; y) y R: y ; y R: y ; y x y; x y x ; x sin (y) 5. f (x; y) e x+y 6. f (x; y) 9 x y R: y ; y R: x ; x 3 x ; x y y x; y x The following regions re unbounded. Sketch the region R nd determine its type. Then nd the volume of the solid under z f (x; y) nd over the given region f (x; y) x y 8. f (x; y) x +y R: x in (; ) ; y in (; ) R: x ; x 9. f (x; y) x e y 3. f (x; y) R: x in (; ) R: x in (; ) y ; y x y x e x ; y x + e x 3. A regulr cone with height h nd bse with rdius R is positioned so tht its xis is horizontl. Find the re A (x) of verticl cross-section of

12 the cone perpendiculr to the xis s function of x in [; h] : Wht is the volume of regulr cone with height h nd bse with rdius R? 3. A hemisphere with rdius R is positioned so tht its xis is horizontl. Find the re A (x) of verticl cross-section of the cone perpendiculr to the xis s function of x in [; R] : Wht is the volume of hemisphere with rdius R? 33. A regulr pyrmid hs height h nd squre bse with ech side length s: It is positioned s shown in the gure below:

13 Find the re A (x) of cross-section t x. Wht is the volume of the pyrmid? 34. The Gret Pyrmid is 48 tll nd hs squre bse which is 756 wide on ech side. Wht is the volume of the Gret Pyrmid? (hint: see problem 33). 35. Explin why the re of type I region cn be written in the form A Z b Z h(x) g(x) dydx 36. Explin why the re of type II region cn be written in the form A Z d Z q(y) c p(y) dxdy 37. Explin why if ; b; c; nd d re ll constnt, then Z b Z d c f (x; y) dydx Z d Z b c f (x; y) dxdy when both iterted integrls exist. 38. Show tht if ; b; c; nd d re constnt, then Z b Z " d Z # " b Z # d f (x) g (y) dydx f (x) dx g (y) dy c 39. Use properties of the integrl to show tht p(x) [f (x; y) + g (x; y)] dy dx p(x) c f (x; y) dy dx+ p(x) g (x; y) dy dx 3

14 4. Use properties of the integrl to show tht [f (x; y) + g (x; y)] dy dx f (x; y) dy dx+ p(x) p(x) p(x) g (x; y) dy dx 4. Show tht if f is di erentible on (; b), then for ll c in (; b) we hve f (c) (b ) + Z b f (x) dx Z b Z x c f (u) dudx 4. Show tht if f is di erentible nd if f () ; then Z b f (x) dx Z b Z f (ux) dudx 4