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a8_soln

# a8_soln - Math 237 1 Evaluate the following integrals a D...

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Unformatted text preview: Math 237 1. Evaluate the following integrals. a) D Assignment 8 Solutions xy 2 dA where D is the region bounded by y = x, y = 2x and x = 3. Solution: From the diagram we see that it is best to draw the rectangles vertically. This gives x ≤ y ≤ 2x and then we have to integrate over all possible x values: 0 ≤ x ≤ 3. Thus we get 3 2x xy 2 dA = D 0 3 x xy 2 dy dx x 0 3 = = 0 13 y 3 2x dx x 74 x dx 3 3 = 75 x 15 = 0 567 5 b) D sin(x + y ) dA where D is the triangular region with vertices (0, 0), (π, 0), and ( π , π ). 22 Solution: From the diagram we see that we want to draw rectanges horizontally. The equation of the two diaganal lies are y = x and y = −x + π . Thus we have y ≤ x ≤ −y + π and then 0 ≤ y ≤ π . Hence, we get 2 π /2 −y +π y π /2 sin(x + y ) dA = D 0 sin(x + y ) dx dy −y +π y = 0 π /2 − cos(x + y ) dy = 0 − cos(π ) + cos(2y ) dy 1 sin 2y 2 π /2 =y+ = 0 π 2 2 1 1 c) 0 y sin(x2 ) dx dy . Solution: We observe that we can not integrate in this order so we want to change the order. We see that the region of integration is y ≤ x ≤ 1 and 0 ≤ y ≤ 1. So, the x values are bounded by the lines y = x and x = 1 and y values run from 0 to 1. Drawing this we get the diagram below. Hence, if we now draw the rectangles vertically we ﬁnd that 0 ≤ y ≤ x and 0 ≤ x ≤ 1 which gives 1 0 y 1 1 x sin(x2 ) dx dy = 0 1 0 sin(x2 ) dy dx x sin(x2 ) dx 0 1 = 1 = − cos(x2 ) 2 0 1 1 = − cos 1 + 2 2 d) D x dA where D is the annular region 1 ≤ x2 + y 2 ≤ 2. Solution: Since our region is bounded by the circles x2 + y 2 = 1 and x2 + y 2 = 2 it would probably√ very helpful to use polar coordinates. In polar coordinates the circles are r = 1 be √ and r = 2 and hence we have 1 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π . Also, in polar coordinates we have x = r cos θ and so we get 2π √ 2 x dA = D 0 2π 1 (r cos θ)r dr dθ cos θ 0 2π = = = 2 0 3/2 13 r 3 √ 1 2 dθ 23/2 − 1 cos θ dθ 3 2π −1 sin θ 3 =0 0 3 e) D ex 2 +y 2 dA where D is the region bounded by y = √ 1 − x2 and y = |x|. Solution: We need to do this in polar coordinates. We ﬁrst observe that y ≥ 0 ⇒ r sin θ ≥ 0 ⇒ 0 ≤ θ ≤ π . Then we get y= r sin θ = r sin θ = √ √ 1 − x2 sin2 θ y = |x| 1 − r 2 cos2 θ r sin θ = |r cos θ| r sin θ = | sin θ| r=1 1 3π/4 sin θ = | cos θ| π 3π θ = or θ = 4 4 π 4 since 0 ≤ θ ≤ π . So the region in polar coorinates is 0 ≤ r ≤ 1 and e D x 2 +y 2 dA = 0 π/4 re r2 π dθ dr = 2 1 ≤θ≤ 3π 4 and so rer dr = 0 2 π (e − 1) 4 f) Dxy 4x2 − y 2 dA where Dxy is the region bounded by the lines y = 2x, y = 2x + 2, y = −2x + 1, y = −2x + 3. Solution: The region D is a parallelogram so to integrate over it we would have to split the region into subregions. Instead we will make a change of variables. Let u = y − 2x and v = y + 2x and then the lines become u = 0, u = 2, v = 1 and v = 3. Hence, we get the nice rectangular region 0 ≤ u ≤ 2 and 1 ≤ v ≤ 3 in the uv -plane. The Jacobian of the −2 1 ∂ (u,v) ) 1 mapping is ∂ (x,y) = det = −4 = 0 and so ∂ (x,y) = ∂ (u,v) = −1 = 1 . Moreover, ∂ (u,v 4 4 21 ∂ (x,y ) 2 2 2 2 we see that −uv = −(y − 2x)(y + 2x) = −(y − 4x ) = 4x − y . Hence, since the mapping has continous partial derivatives we get by the change of variables theorem 2 D 3 1 4x2 − y 2 dA = 0 (−uv ) · 2 1 dv du 4 3 =− =− 1 4 u 0 2 12 v 2 du 1 u du 0 = −2 4 g) Dxy x2 dA, where Dxy is bounded by the ellipse 5x2 + 4xy + y 2 = 1. Solution: Completing the square we get (2x + y )2 + x2 = 1 so we let u = 2x + y and v = x. Since u2 + v 2 = (2x + y )2 + x2 = 4x2 + 4xy + y 2 = 1 we have Duv is a circle of radius 1. Also ∂ (u, v ) 21 = −1, = 10 ∂ (x, y ) so r 2 sin θ and we get 2π 1 ∂ (x,y ) ∂ (u,v) 2 = 1. Using polar coordinates we have u = r cos θ and v = r sin θ so x2 = v 2 = x dA = Dxy 0 0 2π 2 r 3 sin2 θ dr dθ 1 = 4 1 = 4 π = 4 sin2 θ dθ 0 θ1 − sin 2θ 24 2π 0 ...
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