#$&* course MTH272 Sorry there are going to be some back to back post. If your solution to a stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `a*& If the curve is revolved about the x axis the radius of the circle at position x will be the y value radius = `sqrt(4 - x^2). The cross-section at that position will therefore be pi * radius^2 = pi ( sqrt(4 - x^2) ) ^ 2 = pi ( 4 - x^2). The volume of a 'slice' of thickeness `dx will therefore be approximately pi ( 4 - x^2 ) * `dx, which leads us to the integral int(pi ( 4 - x^2) dx, x from 0 to 2). An antiderivative of 4 - x^2 = 4 x - x^3 / 3. Between x = 0 and x = 2 the value of this antiderivative changes by 4 * 2 - 2^3 / 3 = 16 / 3, which is the integral of 4 - x^2 from x = 0 to x = 2. Multiplying by pi to get the desired volume integral we obtain volume = pi ( 16/3) = 16 pi / 3. *&*& DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):NA ------------------------------------------------ Self-critique Rating: *********************************************************** ********************************************* Question: `q5.7.16 (was 5.7.16) vol of solid of rev abt x axis region bounded by y = x^2, y = x - x^2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: bounded by y = x^2, y = x - x^2 we have 2 separate cylinders with y = x - x^2 being the outer part of the cylinder and y = x^2 being an inner cyinder with the volume being the portion between the two. x - x^2=0 x=x^2 combined 2x^2-x=x(2x-1) set to 0 to find x x(2x-1) 2x-1=0 2x=1 x=1/2 x=0 x=1/2 set to 0 to find x pi [(x - x^2)^2]-(x^2) integrate pi{(2x^5)-(5x^4)}/10 pi{(2(1/2)^5)-(5(1/2)^4)}/10 pi(.0625-.3125)/10 pi-.25/10= -.7854/10= -.0785 confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): generally it is exactly the same answer I probably worked it out a little far ------------------------------------------------ Self-critique Rating: *************************************************** ********************************************* Question: `q5.7.18 (was 5.7.18) vol of solid of rev abt y axis y = `sqrt(16-x^2), y = 0, 0
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Given Solution: 2 `a 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):NA ------------------------------------------------ Self-critique Rating: ********************************************************************* ********************************************* Question: `q5.7.31 (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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1/8 x^2 `sqrt(2-x) Using pi*r^2 to calculate the area pi(1/64) [(x^2 `sqrt(2-x)]^2 pi(1/64) (x^4)(2-x) y=0 yields 2 answers (0, 2) 2x^4-x^5 [(2x^5)/5-x^6/6] find antiderivative pi(1/64) [(2(2)^5)/5-(2^6)/6]- [(2*0^5)/5-(0^6)/6] plugging in the 2 points where y=0 (0,2) into the antiderivative to come up with the volume of the solid. pi(1/64) [2*(32/5)]-(64/6)- 0= pi(1/64) [2*(192/30)]-(320/30)= pi(1/64) [384/30)]-(320/30)= (pi/64)(64/30)=pi/30=.105 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):NA ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!