Mathematics, 20.07.2019 08:00, jelopez4

Write the integral in one variable to find the volume of the solid obtained by rotating the first quadrant region bounded by y=0.5x^2 and y=x about the line x=5 note from my teacher: "this has to be a dy not dx for revolving about x=5. also the area is \pi r^2 dy

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If we're revolving around a vertical line, we'll have to integrate with respect to y to use discs method of volumes.in order to integrate with respect to y, we need the curve equations above written in such a way that x is a function of y. currently, with y isolated, the curves are given as y as a function of x.we can graph these out first and then deal with the labeling later. y = 0.5x² is a parabola vertically compressed by a factor of 0.5 or 1/2. y = x is just a line with a slope of 1 and y-intercept of 0. the two graphs intersect at (2,2) and (0,0).converting y = 0.5x² so that x is a function of y, we have to isolate x.note that 0.5 = 1/2. so y = (1/2)x². if we multiply both sides of the equation by 2, we get $2y = x^2$ if we square root both sides, we get $x = \pm \sqrt{2y}$ we take the positive square root because we are in the first quadrant, which involves positive x-coordinates. $x = \sqrt{2y}$ thus the equation $x = \sqrt{2y}$ is equivalent to $y = \tfrac{1}{2}x^2,\ x \ge 0$ .converting y = x so that it has x as a function of y, it's just x = y.to find the volume of the solid that is obatined via rotating this region that forms, we use discs. discs method when integrating with respect to y is $\displaystyle v = \int_a^b \left[ \pi(\text{outer radius})^2 - \pi(\text{inner radius})^2 \right] dy$ the limits of integration will go from 0 to 2 since that is where region starts and stops in terms of y-coordinates. to find the outer radius, we draw a line perpendicular to our revolving line. when we reach our region, we keep going until we hit the back boundary of the region, which is the line x = y for this question. so the outer radius would be the length of that line which is the distance between the revolving line and the curve x = y. $\text{outer radius} = 5 - y$ the inner radius we do the same, except we stop when we touch the front boundary of the curve, which is x = √(2y) in this case. $\text{inner radius} = 5 - \sqrt{2y}$ therefore, the integral is $\displaystyle v = \int_0^2 \left[ \pi(5 - y)^2 - \pi(5-\sqrt{2y})^2 \right] dy$ the question doesn't really say to evaluate this, but if you need this evaluated, you can use a calculator and arrive at the answer of $v = 16\pi/3 \approx 16.755$

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Write the integral in one variable to find the volume of the solid obtained by rotating the first qu...