Calculus Calculus Math Forum

 February 15th, 2011, 12:05 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Please Help......Volume Disk 1. You can't just switch x and y, this results in expressions that are not equivalent to the original, you need to solve for y: $x=\sqrt{y}\:\therefore\:y=x^2$ for non-negative x $x=\frac{y}{4}\:\therefore\:y=4x$ Equating these two expressions for y will determine the limits of integration: $x^2=4x$ $x(x-4)=0$ thus x=0,4 Now, observing $4x>x^2$ on (0,4) we have, using the washer method: $V=\pi\int_0\,^4(4x)^2-$$x^2$$^2\,dx$ $V=\pi\int_0\,^416x^2-x^4\,dx=\pi$\frac{16}{3}x^3-\frac{1}{5}x^5$_0^4=\pi$$\frac{16}{3}4^3-\frac{1}{5}4^5$$=\frac{2048\pi}{15}\approx428.932$ 2. Here the thickness of each disk is dy and the radius is 9 - x = 9 - y² and at x = 9, y = 3 giving: $V=\pi\int_0\,^3$$9-y^2$$^2\,dy=\pi\int_0\,^381-18y^2+y^4\,dy$ $V=\pi$81y-6y^3+\frac{1}{5}y^5$_0^3=\pi$$81\cdot3-6\cdot3^3+\frac{1}{5}3^5$$=\frac{648\pi}{5}\approx 407.150$ 3. Here, we have the thickness of each washer is dx, the outer radius is $\sqrt{x}+1$ and the inner radius is $x+1$ and we integrate from x = 0 to x = 1. $V=\pi\int_0\,^1$$\sqrt{x}+1$$^2-$$x+1$$^2\,dx=\pi\int_0\,^1-x^2-x+2\sqrt{x}\,dx$ $V=\pi$-\frac{1}{3}x^3-\frac{1}{2}x^2+\frac{4}{3}x^{\frac{3}{2}}$_0^1=\pi$$-\frac{1}{3}-\frac{1}{2}+\frac{4}{3}$$$ $V=\frac{\pi}{2}\approx1.571$
 February 20th, 2011, 07:11 PM #3 Newbie   Joined: Feb 2011 Posts: 2 Thanks: 0 Re: Please Help......Volume Disk I finially understand washer's method, thx a lot :P You explained the problems than my teacher, lol.

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# how to revolve around x=9

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