#$&* course Mth 277 December 2 around 3pm. 12.4*********************************************
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Given Solution: The triangular region is in the x y plane and is bounded by x axis, the y axis and the line y = 1 - x. It can therefore be described by 0 <= x <= 1, 0 <= y <= 1 - x. Quick solution: We integrate sqrt(1 + f_x^2 + f_y^2) over the region. For f(x, y) = 2 y, we have sqrt(1 + f_x^2 + f_y^2) = sqrt( 1 + 4 y^2) so our integral is INT(INT(sqrt(1 + 4 y^2) dy, 0, 1 - x), dx, 0, 1), which is approximately .63. This makes sense, because the area of the triangular region is .5, and the parabolic region makes a relatively modest angle with vertical above most of the triangle. Above the point (x, y), a vector normal to the surface z = y^2 is the cross product of i + f_x k and j + f_y k, where f(x, y) = y^2. This gives us the normal vector i X (j + 2 y k) = k - 2 y i. Alternatively the surface z = f(x, y) is a level surface of the function F(x,y,z) = z - f(x, y). Since the gradient of this function is normal to the level surface, the vector del F(x, y, z) is normal to the surface. In this case the surface is a level curve of F(x, y, z) = z - f(x, y) = z - y^2, so F_x = 0, F_y = -2 y and F_z = 1. The gradient is therefore 0 i - 2 y j + k., the same as that found previously. In general the gradient of z - f(x, y) is -f_x i - f_y j + k, and the cosine of its angle with the k vector is 1 / sqrt( 1 + f_x^2 + f_y^2 ). The unit normal to the xy plane is k, so the cosine of the angle between the normal to the surface and the normal to the xy plane is cos(theta) = ( k - 2 y i ) dot k / | k - 2 y i | = 1 / sqrt( 1 + (2y)^2 ) = 1 / sqrt( 1 + 4 y^2 ). The portion of the our surface which lies above an area increment `dA with dimensions `dx by `dy, located at (x_hat, y_hat) has an area that is equal to `dA, if the surface is parallel to the x y plane, and an area greater than `dA, if the surface is not parallel to the x y plane. The ratio of the areas is 1 / cos(theta) = sec(theta), where theta is the angle between the normal to the surface and the normal to the plane. The cosine of that angle was found above to be 1 / sqrt( 1 + 4 y^2 ). The reciprocal is sec(theta) = sqrt( 1 + 4 y^2 ). For the sample point (x_hat, y_hat) we have sec(theta) = sqrt( 1 + 4 y_hat^2 ). So the surface area corresponding to our area increment is sqrt( 1 + 4 y_hat^2 ) `dA = sqrt( 1 + 4 y_hat^2 ) `dy `dx. Our Riemann sum leads us to the integral int ( int ( sqrt(1 + 4 y^2) dy, 0, 1 - x) dx, 0, 1) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK, except I got (5/6)=0.83 and you got 0.63??!?
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Given Solution: The sphere intersects the plane when x^2 + y^2 + 4^2 = 36 so that x^2 + y^2 = 20. This describes a circle of radius 2 sqrt(5) centered at the origin. This circle can be described by -2 sqrt(5) <= x <= 2 sqrt(5) -sqrt(20 - x^2) <= y <= sqrt(20 - x^2) A vector normal to the sphere at point (x, y, z), obtained using the gradient, is 2 x i + 2 y j + 2 z k. This should be no surprise, since the radial vector x i + y j + z k is normal to the surface at (x, y, z); the first of these is just double the first. Using either we will find that cos(theta) = z / sqrt(x^2 + y^2 + z^2) = sqrt(36 - x^2 - y^2) / 6, so we will integrate 6 / sqrt( 36 - x^2 - y^2) over our circle.. Alternatively we can solve the equation of the sphere for z, obtaining z = f(x, y) = sqrt( 36 - x^2 - y^2 ). In this case f_x = - x / sqrt( 36 - x^2 - y^2), f_y = - y / sqrt( 36 - x^2 - y^2), and sqrt( 1 + f_x^2 + f_y^2) = 6 / sqrt( 36 - x^2 - y^2). We integrate the expression sqrt( 1 + f_x^2 + f_y^2) = 6 / sqrt( 36 - x^2 - y^2) over the region, obtaining int(int(6 / sqrt( 36 - x^2 - y^2) dy, -sqrt(20 - x^2) , sqrt(20 - x^2) ) dx, -2 sqrt(5), 2 sqrt(5) ). The result is 24 pi. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK, now I see. The sqrt (1+fx^2+fy^2)=6/sqrt(36-x^2-y^2) and then integrating over the region. Got it! ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q003. Find the surface area of the portion of the cylinder x^2 + z^2 = 9 which lies above the triangle with vertices (0,0,0), (1,1,0), (1,0,0). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: **** z^2=(9-x^2) z=sqrt (9-x^2) limits of integration: [0<=x<=1], [0<=y<=1-x] [Int Int ((sqrt 9)/(sqrt (9-x^2) dy, 0, 1-x)dx, 0, 1] Solving this double integral, I got: 3*[sin^(-1)*(1/3)+2*sqrt(2)-3] #$&* confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The cylinder has gradient vector 2 x i + 2 z k, so cos(theta) = 2 z / 2 sqrt( x^2 + z^2) = z / sqrt( x^2 + z^2) and sec(theta) = sqrt( x^2 + z^2 ) / 2 z. Since z = sqrt(9 - x^2) this simplifies to sqrt( 9 ) / ( sqrt( 9 - x^2) ). This is the function we integrate over the region. Alternatively z = f(x,y) = sqrt( 9 - x^2) so f_x = -x / sqrt( 9 - x^2), f_y = 0 and sqrt( 1 + f_x^2 + f_y^2 ) = sqrt( 1 + x^2 / (9 - x^2) ) = sqrt( 9 / (9 - x^2) ) = 3 / sqrt( 9 - x^2 ). Either way this function is integrated over the region 0 < x <= 1, 0 <= y <= 1 - x. The integral is int( int( 3 / sqrt( 9 - x^2) dy, 0, 1-x) dx, 0, 1) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: