# using the washer method to find the volume of an object

When finding the volume of a solid generated by revolving $y=2x+8$ and $y=2x^2+1$, about the $x$ axis, I will first set them equal to each other to find the bounds for my integral. Doing this I get $2.44$ and $-1.44$, which $a$ = $2.44$ and $b$ = $-1.44$ for my bounds. Now I have

$\pi$$\int^{2.44}_{-1.44} (2x+8)^2\,dx$-$(2x^2+1)^2dx$ in the form of $V=\int^{a}_{b}\pi(R(x)^2-r(x)^2)dx$. Now I will take the anti derivative of each function and arrive at

$\pi(\frac{4x^3}{3}+\frac{32x^2}{2}+64x) - (\frac{4x^5}{5}+\frac{4x^3}{3}+{x})^{2.44}_{-1.44}$ Where do I go from here? Do I put bounds $a$ and $b$ into $R(x)$ - $r(x)$ or do I only do $R(a)$ - $R(b)$ ?

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You could start by describing the object whose volume you are trying to calculate. You give two functions, so probably you want the area between them rotated around some axis. At the end we get $R(x), r(x), a, b$, none of which have been defined. – Ross Millikan Jan 28 '13 at 1:06
Just edited the main post. – Kyle H Jan 28 '13 at 3:12
Ah man, so many mistakes I overlooked. I am editing this right now. – Kyle H Jan 28 '13 at 4:28
Ok, that last edit should do it. – Kyle H Jan 28 '13 at 4:34
You have two integrals which are identical. Please ask yourself which function is bigger, to make sure your answer will be positive. Sketch a graph of your functions so you can visualize what a typical washer element you are rotating looks like. You should then edit the question. – Barbara Osofsky Jan 28 '13 at 4:52

Your endpoints appear reasonable approximations so I will use them, even though you could get exact values using the quadratic formula. You might check with your teacher about that.

The washer method says that the volume is equal to $\pi \int_{-1.44}^{2.44} \left(2x^2 +1 \right)^2 dx - \pi \int_{-1.44}^{2.44} \left(2x+8 \right)^2 dx$. Each of those definite integrals has a numerical value which is computed as you indicate by taking antiderivatives and evaluating them at the endpoints.

The first integral gives the volume of the solid obtained by rotating only the parabola. The integrand has antiderivative which you calculate correctly to get that volume is $\pi \left({{4x^5}\over 5} + ({{4x^3}\over 3}+x\right)\bigg\vert _{-1.44}^{2.44}$ which is a number obtained by plugging the upper limit 2.44 into the expression to the left of the $\big\vert$ and then subtracting the result of the number obtained by plugging the lower limit into the expression and then subtracting it from the previous number. This will give you a number with no $R$, $r$, $x$, $a$, or $b$ as the volume swept out by the parabola.

Now do the same thing to get the volume of the solid generated by rotating the line. The absolute value of the difference of these two numbers will give you the answer you are seeking.

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Given the form $\int_b^a \pi (R(x)-r(x)) \; dx$ (note-you forgot the $dx$), you can split it into $\int_b^a \pi R(x) dx - \int_b^a \pi r(x)$ then integrate each term as you have. If $S=\int R(x), s=\int r(x)$, then you have $\pi(S(a)-S(b)-s(a)+s(b))$. Your expression just before"Where do I ..." is fine aside from mismatched parentheses and the fact that the leading $\pi$ should multiply the second trinomial as well as the first. Just plug in $x=2.44$ and $x=-1.44$ and subtract.

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It appears misleading to use the letter R or r to represent the radius squared rather than the radius. Without a picture there are other problems. The computed antiderivatives are completely ignored when the final question is asked. – Barbara Osofsky Jan 28 '13 at 5:45
@Kyle The above comment no longer applies now that the question has been edited. Good job Kyle. And you even caught the sign flip! – Barbara Osofsky Jan 29 '13 at 4:12