course mth 272 ՁQĕܯyassignment #013
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21:54:46 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 --> 3.14 (4-x^2) = some of volume the integral from that is 3.14 ( 4 - x^2) dx at 0 and 2
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21:55:33 *&*& 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 --> and then find the derivative. which is 4 x - x^3 / 3 solved for 16/3 then multiply 3.14 to equal 16 pi/3
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11:14:59 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 --> x=0 and x= 12 because x-x^2 = x^2 and intersects area of the circles: 3.14((x-x^2)^2-x^2) so - 3.14/ 40 or -pi/40
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11:15:08 ** 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 --> ok
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11:19:24 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 --> x = square root(16 - y^2) is the radius ...
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11:20:14 ** 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 --> then integrate pi x^2 = pi ( sqrt(16 - y^2) ) ^2 = pi ( 16 - y^2) from y = 0 to y = 4 and then do the antiderivative An antiderivative of pi ( 16 - y^2) with respect to y is pi ( 16 y - y^3 / 3) getting int( pi ( 16 - y^2), y, 0, 4) = pi ( 16 * 4 - 4^3 / 3) - pi ( 16 * 0 - 0^3 / 3) = 128 pi / 3
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11:23:34 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 --> cross sectional area is 3.14/64 * ( 2 x^4 - x^5) so volume = int(3.14/64 * ( 2 x^4 - x^5), x, 0, 2)
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11:24:10 ** 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 --> get the antiderivative : 2 x^4 - x^5 is 2 x^5 / 5 - x^6 / 6 hence 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
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