#$&* course Mth 277 4/15 1 Question: `q001. Find the surface area of the portion of the surface z = y^2 that lies over the triangular region in the plane with vertices (0,0,0), (1,0,0) and (0,1,0).YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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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, 1, 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) STUDENT QUESTION OK, except I got (5/6)=0.83 and you got 0.63??!? INSTRUCTOR RESPONSE .63 is correct to 2 significant figures. 5/6 can't be right. The integral is way too complicated for a rational-number solution. Square roots, at the very least, would be required. An antiderivative for the y integral is ln(sqrt(4y^2 + 1) + 2y)/4 + y* sqrt(4 y^2 + 1)/2. You get this by starting with the trig substitution y = tan(theta) / 4. You use the identity 1 + tan^2(theta) = sec^2(theta), and y ' = tan(theta) sec(theta) / 4 so dy = 1/4 tan(theta) sec(theta) dTheta. Fairly straightforward, if you're up on trig substitution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot to square f_y, which is the reason for my incorrect answer. ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. Find the surface area of the portion of the sphere x^2 + y^2 + z^2 = 36 that lies above the plane z = 4. 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I didn’t look at solution, First we calculate Partial derivatives f_x = 2x f_y = 2y So our surface integral is Int(Int( `sqrt(f_x^2 + f_y^2 + 1 `dA) = Int(Int( 4x^2 + 4x^2 + 1 `dA)
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ 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: We first find the fn of the surface area which is z = `sqrt( x^2 + 9), now we would find partial dervatives I believe f_x = f(g(h))’ = f’(g(h)*g’(h) = (1/(2(x^2+9)^(1/2)))*(2x) = x/(x^2 + 9)^(1/2) f_y = 0 So our surface integral is Int(Int( `sqrt(f_x^2 + f_y^2 + 1 `dA) = Int(Int(`sqrt([x/(x^2 + 9)^(1/2)]^2 + 1 dy, 0,1-x) x,0,1) Int(`sqrt([x^2/(x^2 + 9) + 1) dy) = `sqrt([x^2/(x^2 + 9) + 1)*Int( 1 dy) = `sqrt(x^2/(x^2 + 9) + 1)*y,evaluating `sqrt(x^2/(x^2 + 9) + 1)*(1 - x) Int(`sqrt(x^2/(x^2 + 9) + 1)*(1 - x)dx) 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): I finally see, where I was to have z =`sqrt(9-x^2) I had z = x^2 + 9 Int(3 / sqrt( 9 - x^2) dy) = y*(3 / sqrt( 9 - x^2)), evaluated (1-x)* (3 / sqrt( 9 - x^2) = -3*(x - 1)/(`sqrt(9 - x^2) Int(-3*(x - 1)/(`sqrt(9 - x^2) dx), using Trig substitution x = 3sin(`theta), dx = 3cos(`theta) `dtheta, with 9-x^2 = 9cos(`theta)^2. So… Int(-3*(x - 1)/(`sqrt(9 - x^2) = Int([-3*(3cos(`theta - 1)/(`sqrt(9cos(`theta)^2)]*3cos(`theta) `dtheta) = Int((3cos(`theta)*-3*(3sin(`theta )- 1))/(3cos(`theta)) `dtheta) = Int((-3*(3sin(`theta )- 1) `dtheta) = 3(3cos(`theta) + `theta), where x = 3sin(`theta) 3(3cos(`theta) + `theta) = 3sin^-1(x/3) + 3`sqrt(9-x^2), so that Int(-3*(x - 1)/(`sqrt(9 - x^2) = 3(sin^-1(x/3) + `sqrt(9 - x^2)), evaluated 3(sin^-1(1/3) + `sqrt(9 - 1^2)) - 3(sin^-1(0/3) + `sqrt(9 - 0^2)) = 3sin^-1(1/3) + 6*`sqrt(2) - 9 ????Think this is the correct solution???????????? ------------------------------------------------ Self-critique rating: