Find the volume of the solid obtained by rotating the region bounded by the give

Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about a specific point. Sketch the region, the solid, and a typical disk or washer. y=x^2, x=y^2, about y=1. So i took the integral from 0 to 1 A(x)dx=pi(sqrt(x)^2-(x^2)^2)=0 to 1 of pi*(x-x^4)dx. I took the antiderivative of that and I got the integral from 0 to 1 of x^2/2-x^5/5. I evaluated x at the limits and got 3pi/10. The answer in the book however is 11pi/30. How do i solve this problem? First and best rating to the one who gives the most detailed and correct and answer.

Explanation / Answer

x = y - y^2, x = 0; about the y-axis Washer method called for here. Each washer will have an outer radius R = x = y - y² and an inner radius r = 0 and a thickness of dy, so V = ?[0,1] p(R² - r²) dy = p?[0,1] (y - y²)² dy = p?[0,1] (y² - 2y³ + y^4) dy V = p((1/3)y³ - ½y^4 + (1/5)y^5) |[0,1] = p(1/3 - 1/2 + 1/5) = (p/30)(10 - 15 + 6) V = p/30 y = sec x, y = 1, x = -1, x = 1; about the x-axis Here, too, washer method is called for. Each washer will have an outer radius R = sec x and an inner radius r = 1 and a thickness dy, so V = ?[a,b] p(R² - r²) = p?[-1,1] (sec²x - 1) dx = p(tan(x) - x) |[-1,1] Take it from here. y = x2/3, x = 1, y = 0; about the y-axis Shells or washers work here. Each shell will have a radius R = x and a height h = y = x^(2/3) and a thickness of dx, so V = ?[a,b] 2pRh dx = 2p?[0,1] x*x^(2/3) dx = 2p?[0,1] x^(5/3) dx V = 2p(3/8)x^(8/3) |[0,1] = 3p/4 == OR == Each washer will have an outer radius R = 1 and an inner radius r = x = y^(3/2) and a thickness dy, so V = ?[a,b] p(R² - r²) dy = p?[0,1] (1 - y³) dy = p(y - (1/4)y^4) |[0,1] V = p(1 - 1/4 - 0 + 0) = 3p/4

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