I need help following volumes: 1) Find the volume V of the solid obtained by rot

Question

I need help following volumes: 1) Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 3e^?x, y = 3, x = 2; about y = 6 2)Find the volume V of the described solid S. The base of S is an elliptical region with boundary curve 4x^2 + 25y^2 = 100. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base. 3)Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 8x^3, y = 0, x = 1; about x = 2 4)Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 1/49^x2, x = 5, y = 0; about the y-axis 5)Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = ln 7x, y = 2, y = 3, x = 0; about the y-axis 6)Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 8 sin x, y = 8 cos x, 0 ? x ? ?/4; about y = ?1 Please show a step by step so I know how to do it and whats going on. Thanks so much for your help I promise to rate back!!!!

Explanation / Answer

1 ans) he curves y = 3e^(-x) and y = 6 intersect at the point (0, 3=), and both continue past x = 5. For any x between 0 and 5, the cross-section of the solid is an annulus having these parameters: exterior radius = 6 - 3e^(-x) interior radius = 3 area of annulus = p{[6 - 3e^(-x)]² - 3²} volume of revolution = p?{[6 - 3e^(-x)]² - 3²}dx, with limits 0 and 5 = p?[27 - 36e^(-x) + 9e^(-2x)]dx = p[27x + 36e^(-x) - (9/2)e^(-2x)], still with limits 0 and 5 = p[135 + 36e^(-5) - (9/2)e^(-10) - 36 + 9/2] = p[207/2 + 36e^(-5) - (9/2)e^(-10)] = (9/2)p[23 + 8e^(-5) - e^(-10)] 2 ans) "Cross-sections perpendicular to the x-axis" indicates that we're going to be measuring the hypotenuse of each triangle in the y-direction. Then y² = (1/81)(324 - 4x²) We're told that each triangle's hypotenuse lies in the base of the ellipse, therefore each hypotenuse = 2y in length (above and below the x-axis). Use geometry to get the height of each triangle: Bisecting each one vertically (that is, via a vertical plane including the x-axis) we see "half" triangles with base y and height h, where tan45º = 1 = opp/adj = h / y, so h = y. Therefore the area of each complete (unbisected) triangle A = ½bh = ½(2y)y = y² Sum these triangles using integration: V = ?[-9,9] y² dx = ?[-9,9] (1/81)(324 - 4x²) dx = (1/81)(324x - (4/3)x³) |[-9,9] V = (1/81)(2916 - 972 - (-2916 + 972)) = (1/81)(3888) = 48

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