#$&* course MTH 272 10/16 about 3:45pm 013. `query 13
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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):See question in solution. I did not see anywhere stating to evaluate from 0 to 2. Also, I did not see any review questions for this section 5.7 nor do I have a section 5.7 in my book to be able to reference to another problem. Although, I did work the problem for both intervals but I only showed from 0 to 2 on this assignment. ------------------------------------------------ Self-critique Rating:3
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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):My answer varies from your answer. See question in the solution.
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Given Solution: `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):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* 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: Setting y=0 and solving for x you get x=2 and x=0. This is our interval. A=’pi*r^2=’pi * (1/8 * x^2 * ‘sqrt(2-x))^2 multiply, simplify, and factor out ‘pi/64 A= (‘pi/64)*(2x^4-x^5) V= int((‘pi/64)*(2x^4-x^5) dx, x, 0, 2) factor out ‘pi/64 V=’pi/64 * int. ((2x^4-x^5) dx, x, 0, 2) V=(‘pi/64) * [(2x^5/5) - (x^6/6)] from 0 to 2 V=(‘pi/64) * [((2*2^5)/5) - (2^6/6) - 0] V=(‘pi/64) * [(64/5) - (64/6)] = (‘pi/64) * (32/15) = ‘pi/30 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!