#$&* course Mth 277 Question: `q001. Compute the iterated triple integral Int( Int( Int(x^2*y sin(xyz) dz, 0,1) dy,0,1) dx,0, pi)
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Given Solution: The inner integrand is x^2 y sin(x y z) with respect to z. Antiderivative is -x^2 y * (cos(x y z) / x y) = x cos(x y z). Between z = 0 and z = 1 this changes from -x cos(0) to -x cos(x y), a change of x - x cos(x y). The middle integrand is therefore x - x cos(x y), integrated with respect to y. Antiderivative is - x sin(x y) / x + x y / 2 = -sin(x y) + xy. Between y = 0 and y = 1 this changes from - sin(0) + 0 to -sin(x) + x, a change of -sin(x) + x. The outer integrand is therefore - sin(x) + x = cos(x) + x^2 / 2. The change in the value of the antiderivative between x = 0 and x = pi is thus cos(pi) + pi^2 / 2 - cos(1) = -2 + pi^2 / 2. In standard form the integral is written &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002. Evaluate the triple integral of xz + 2yx over D with respect to V where D is the box 2<= x <= 4, 1 <= y <= 3, -1 <= z <= 1 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2<= x <= 4, 1 <= y <= 3, -1 <= z <= 1 xz + 2yx Int (xz+2yx dx) 2 to 4 =x^2(2y+z)/2 2 to 4 =6(2y+z) Int (6(2y+z) dy) 1 to 3 =6y(y+z) 1 to 3 = 12(z+4) Int (12(z+4) dz) -1 to 1 =6z(z+8) -1 to 1 =98 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 got 98 instead of 96 so I must of made a small mistake somewhere, but I believe my work is satisfactory.
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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):OK ------------------------------------------------ Self-critique rating:OK ********************************************* 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: Int(Int(Int(f(x,y,z) dz, 0, 1-2x) dy, 0, 1-4x^2) dx, 0, 1/2) to dy dx dz z = 1-2x y = 1-4x^2 x = z/2 -1 x = `sqrt((y-1)/4) 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):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q005. Use triple integration to find the volume of the following solids. A sphere of radius r. An ellipsoid with equation x^2/a^2 + y^2/b^2 + z^2/c^2 = 1 (a,b,c > 0). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Volume of: A sphere of radius r. An ellipsoid with equation x^2/a^2 + y^2/b^2 + z^2/c^2 = 1 (a,b,c > 0) -1<= x <= 1 -sqrt(1 - x^2) <= y <= sqrt(1 - x^2) z = +-sqrt( 1 - x^2 - y^2) -1<= x <= 1 -sqrt(1 - x^2) <= y <= sqrt(1 - x^2)
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There was not a solution so I am unsure if I am on the right track. ------------------------------------------------ Self-critique rating:OK"
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