course mth 272 ???????b???K??assignment #013013. `query 13 FINISH THIS ONE UP AND POST
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12:35:39 5.7.4 (was 5.7.4 vol of solid of rev abt x axis, y = `sqrt(4-x^2) What is the volume of the solid?
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RESPONSE --> use disk method integral of pi `sqrt((4-x^2)^2)^2 dx at x = -2 and x = 2 expand integrand not sure how to do this with an sqrt raised to a power do they cancel? which would leave (4-x^2) antiderivative = pi[x^3/3] apply fundamental theorem pi/16 =.196 confidence assessment: 1
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12:40:09 *& If the curve is revolved about the x axis the radius of the circle at position x will be the y value radius = `sqrt(4 - x^2). The cross-section at that position will therefore be pi * radius^2 = pi ( sqrt(4 - x^2) ) ^ 2 = pi ( 4 - x^2). The volume of a 'slice' of thickeness `dx will therefore be approximately pi ( 4 - x^2 ) * `dx, which leads us to the integral int(pi ( 4 - x^2) dx, x from 0 to 2). An antiderivative of 4 - x^2 = 4 x - x^3 / 3. Between x = 0 and x = 2 the value of this antiderivative changes by 4 * 2 - 2^3 / 3 = 16 / 3, which is the integral of 4 - x^2 from x = 0 to x = 2. Multiplying by pi to get the desired volume integral we obtain volume = pi ( 16/3) = 16 pi / 3. *&*& DER
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RESPONSE --> i messed up on the antiderivative and took the derivative. should be antiderivative of 4 - x^2 = 4 x - x^3 / 3 beteween x = 0 and x = 2. 4 * 2 - 2^3 / 3 = 16 / 3 pi(16/3) = 16pi/3 self critique assessment: 2
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12:46:48 5.7.16 (was 5.7.16) vol of solid of rev abt x axis region bounded by y = x^2, y = x - x^2
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RESPONSE --> use washer method find points of intersection of f and g f(x) = g(x) x^2 = x- x^2 2x^2 = x x^2 = x/2 x= +- x/2 set up f(x)^2 - g(x)^2 dx i dont think I have the right point of intersection to finish the problem confidence assessment: 0
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12:52:16 y = x^2 and y = x - x^2 intersect where x^2 = x - x^2, i.e., where 2x^2 - x = 0 or x(2x-1) = 0. This occurs when x = 0 and when x = 1/2. So the region runs from x=0 to x = 1/2. Over this region y = x^2 is less than y = x - x^2. When the region is revolved about the x ais we will get an outer circle of radius x - x^2 and an inner circle of radius x^2. The area between the inner and outer circle is `pi (x-x^2)^2 - `pi(x^2)^2 or `pi ( (x-x^2)^2 - x^2). We integrate this expression from x = 0 to x = 1/2. The result is -`pi/40. ** DER
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RESPONSE --> points of intersection are at x=0 to x = 1/2 Over this region y = x^2 is less than y = x - x^2 This leads to an outer circle f(x)of radius x - x^2 and an inner circle g(x) of radius x^2 The area between the inner and outer circle = pi (x-x^2)^2 - `pi(x^2)^2 integrated from x = 0 to x = 1/2. = - pi/40 self critique assessment: 2
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12:59:49 5.7.18 (was 5.7.18) vol of solid of rev abt y axis y = `sqrt(16-x^2), y = 0, 0
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RESPONSE --> setting `sqrt(16-x^2) = 0 x = +-4 use washer method formula simplifying 16-x^2 dx at x = -4, x= 4 antiderivative = 16x - x^3/3 42.666 - -42.666 = 0 confidence assessment: 1
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13:07:33 At x = 0 we have y = 4, and at x = 4 we have y = 0. So the y limits on the integral run from 0 to 4. At a given y value we have y = sqrt(16 - x^2) so that y^2 = 16 - x^2 and x = sqrt(16 - y^2). At a given y the solid will extend from x = 0 to x = sqrt(16 - y^2), so the radius of the solid will be sqrt(16 - y^2). So we integrate pi x^2 = pi ( sqrt(16 - y^2) ) ^2 = pi ( 16 - y^2) from y = 0 to y = 4. An antiderivative of pi ( 16 - y^2) with respect to y is pi ( 16 y - y^3 / 3). We get int( pi ( 16 - y^2), y, 0, 4) = pi ( 16 * 4 - 4^3 / 3) - pi ( 16 * 0 - 0^3 / 3) = 128 pi / 3. This is the volume of the solid of revolution. ** DER
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RESPONSE --> oh dont set them equal to each other you use the 0
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13:13:53 5.7.31 (was 5.7.30) fuel take solid of rev abt x axis y = 1/8 x^2 `sqrt(2-x) What is the volume of the solid?
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RESPONSE --> use disk method antiderivative = 1/4x * 2x -x^2/2 at x = 0 and x = 2 4pi/2 confidence assessment: 1
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13:20:13 The radius of the solid at position x is 1/8 x^2 sqrt(2-x) so the cross-sectional area at that point is c.s. area = pi * (1/8 x^2 sqrt(2-x) )^2 = pi/64 * x^4(2-x) = pi/64 * ( 2 x^4 - x^5). The volume is therefore vol = int(pi/64 * ( 2 x^4 - x^5), x, 0, 2). An antiderivative of 2 x^4 - x^5 is 2 x^5 / 5 - x^6 / 6 so the integral is pi/64 * [ 2 * 2^5 / 5 - 2^6 / 6 - ( 2 * 0^5 / 5 - 0^6 / 6) ] = pi / 64 * ( 64 / 30) = pi/30. ** DER
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RESPONSE --> i thought the sqrt and the exponent cancelled each other in the entire problem, correct answer though should be pi * (1/8 x^2 sqrt(2-x) )^2 = pi/64 * x^4(2-x) = pi/64 * ( 2 x^4 - x^5) vol = int(pi/64 * ( 2 x^4 - x^5), x, 0, 2) take antiderivative = 2 x^5 / 5 - x^6 / 6 pi/64 * [ 2 * 2^5 / 5 - 2^6 / 6 - ( 2 * 0^5 / 5 - 0^6 / 6) ] = pi / 64 * ( 64 / 30) = pi/30 self critique assessment: 2
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