#$&* course Mth 277 4/18 Question: `q001. Compute the iterated triple integral Int( Int( Int(x^2*y sin(xyz) dz, 0,1) dy,0,1) dx,0, pi)
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Given Solution: The inner integrand is x^2 y sin(x y z) with respect to z. Antiderivative is -x^2 y * (cos(x y z) / x y) = x cos(x y z). Between z = 0 and z = 1 this changes from -x cos(0) to -x cos(x y), a change of x - x cos(x y). The middle integrand is therefore x - x cos(x y), integrated with respect to y. Antiderivative is - x sin(x y) / x + x y / 2 = -sin(x y) + xy. Between y = 0 and y = 1 this changes from - sin(0) + 0 to -sin(x) + x, a change of -sin(x) + x. The outer integrand is therefore - sin(x) + x = cos(x) + x^2 / 2. The change in the value of the antiderivative between x = 0 and x = pi is thus cos(pi) + pi^2 / 2 - cos(1) = -2 + pi^2 / 2. In standard form the integral is written &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. Evaluate the triple integral of xz + 2yx over D with respect to V where D is the box 2<= x <= 4, 1 <= y <= 3, -1 <= z <= 1 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int( Int( Int(xz + 2yx dz, -1,1) dy,1,3) dx,2,4) Int(xz + 2yx dz ) = (x)Int( z dz) + (2xy)*Int(1 dz) (x)Int( z dz) = x(z^2/2) (2xy)*Int(1 dz) = (2xy)*(z) = 2xyz Int( x - xcos(xy) dy) = x(z^2/2) + 2xyz, evaluating [ x(1^2/2) + 2xy*1] - [x(-1^2/2) + 2xy*(-1)] = 4xy Int( Int( 4xy dy,1,3) dx,2, 4) Int( 4xy dy) =4x* Int( y dy) = 4x*(y^2/2), evaluating [ 4x*(3^2/2),] - [4x*(1^2/2),] = 18x - 4x = 14x Int(14x dx,2,4) Int(14x dx) = 14*Int( x dx) = 14*(x^2/2), evaluated [14*(4^2/2) ] - [14*(2^2/2)] = 14*(8 - 2) = 14*6 = 84 Int( Int( Int(xz + 2yx dz, -1,1) dy,1,3) dx,2,4) = 84 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The integral is int(int(int((xz+2yx) dz, -1, 1) dy, 1, 3) dx, 2, 4) The inner integrand is x z + 2 y x, integrated with respect to z, which gives us antiderivative x z^2 / 2 + 2 y x z. Between z = -1 and 1 this antiderivative changes from x / 2 - 2 y x to x / 2 + 2 y x = 4 y x. The middle integrand is therefore 4 y x with respect to y, which gives us antiderivative 2 y^2 x. Between y = 1 and 3 this antiderivative changes from 2 x to 18 x, a change of 16 x. The outer integrand 16 x is easily evaluated between x = 2 and x = 4. Antiderivative is 8 x^2, which changes from 32 to 128, a change of 96. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not divide 4x by 2 to get 2x, so when evaluated instead of 16x I was working with 14x. ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003. Use a triple integral to find the volume of the solid bounded by y = 4 - x^2, z = 0, and z = y. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For y = 4-x^2, we have a parabola with orgin at (0,4) in the xy plane, so y has domain of 4 to -infinity but because we are looking a in the boundary of z=0 and z = y, z then has a max of 4(max value of y) and we are only looking at the parabola from y=4 to x axis(y=0) so our integral is Int(Int(Int( dz,0,4) dy,0,4-x^2) dx,-2,2) Int( dz) = z,evaluated 4 - 0 = 4 Int( 4 dy) = 4y,evaluated 4*(4-x^2) = 16x^2 Int(16 - 4x^2 dx) = 16x - (4/3)x^3,evaluated [32 - (4/3)*8] - [32 -(4/3)8] = 64 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: y = 4 - x^2 is a parabolic cylinder extending above and below the parabola y = 4 - x^2 in the x y plane. This parabola has vertex at (0, 4) and opens in the negative y direction, passing through the x axis at (2, 0) and (-2, 0). The plane z = 0 is just the x y plane, and the plan z = y 'slices' 3-dimensional space at an angle of 45 degrees to the x y plane, 'cutting through' at the x axis. The region of the cylinder above the x y plane and below the z = y plane is finite. The region above the z = y plane and below the x y plane is infinite in extent. We integrate to find the volume of the finite region. In the x y plane the region beneath the cylinder can be described as -2 <= x < = 2, 0 <= y <= 4 - x^2, 0 <= z <= y. Our integral is therefore int(int(int(dz, 0, y) dy, 0, 4 - x^2), dx, -2, 2). Our inner integrand is just 1 with respect to z, with antiderivative z. Between z = 0 and z = y our antiderivative changes by y. The integrand of our middle integral is therefore y, integrated with respect to y, giving us antiderivative y^2 / 2. Between y = 0 and y = 4 - x^2 this antiderivative changes by (4 - x^2)^2 / 2 = (16 - 8 x^2 + x^4) / 2 = 8 - 4 x^2 + x^4/2. The integrand of our outer integral is therefore 8 - 4 x^2 + x^4/2, integrated with respect to x. Our antiderivative is 8 x - 4/3 x^3 + x^5 / 10. Between -2 and 2 this antiderivative changes by 2 * (16 - 4/3 * 8 + 32/10) = 256/15. It might be instructive to contrast this with the double integral which yields the same result. For each area increment `dA with sample point (x_hat, y_hat), the corresponding region of our solid extends from z = 0 to z = y_hat. So the volume of the region is y_hat * `dy * `dx. The resulting Riemann sum approaches the integral int(int(y dy, 0, 4-x^2) dx, -2, 2) = 256/15. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I put z boundary from 0 to 4 but I guess it should have been from 0 to y. I believe that was my only mistake. ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. Change the order of integration of the triple integral Int(Int(Int(f(x,y,z) dz, 0, 1-2x) dy, 0, 1-4x^2) dx, 0, 1/2) to dy dx dz. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Everything is given in terms of x so we have z = 1-2x y = 1-4x^2 Solving for x, respectively, x = z/2 -1 x = `sqrt((y-1)/4) Looking at graph of y = 1-4x^2, where x is from 0 to (1/2) as in original boundary the integral is aimed to find the area within the first Quad and we can see that where x runs from 0 to (1/2) that y runs from 0 to 1 during the interval defined by y = 1-4x^2 and for 0<=x<=(1/2) so now we put together our integral Int(Int(Int( f(x,y,z) dy,0,1) dx,0,`sqrt((y-1)/4)) dz,0, `sqrt((y-1)/4)) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The region can be described by 0 <= x <= 1/2, 0 <= y <= 1 - 4 x^2, 0 <= z <= 1 - 2 x. 0 <= x <= 1/2, 0 <= y <= 1 - 4 x^2 describes the region between the x axis and the parabola y = 1 - 4 x^2. The vertex is (0, 1) and the x intercepts are (+- 1/2, 0). This region can also be described as follows: For any y between 0 and 1, the horizontal line through (0, y) meets the boundary curve at coordinate x, such that y = 1 - 4 x^2. The value of x is therefore x = +- sqrt( (1 - y) / 4) = +- 1/2 sqrt( 1 - y ). The horizontal line is inside the region at the point (0, y) and therefore remains inside the region between these two values of x. The region in the x y plane can therefore be described as 0 <= y <= 1/2, -1/2 sqrt( 1 - y ) <= x <= 1/2 sqrt( 1 - y ). ... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??Why is the defined region in the x y plane described as 0 <= y <= 1/2, -1/2 sqrt( 1 - y ) <= x <= 1/2 sqrt( 1 - y ) and not 0<=y<=1?????
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):