Question

Find the volume of the solid formed by rotating the region inside the first quadrant enclosed by y=x^3 and y=9x about the x-axis.
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Answer 1

First solve x³=9x

$x^3-9x=0 \\x(x^2-9)=0 \\x(x-3)(x+3)=0 \\x=0, x=3,x=-3$

We can ignore x = -3 because it is not in the first quadrant.

So our integral is going to go from x=0 to x=3.

Now we can use the formula
$V=\int_{a}^{b}\pi f(x)^2dx$
$\int\limits_{0}^{3}(\pi (9x)^2-\pi(x^3)^2)dx \\\int\limits_{0}^{3}(\pi (9x)^2-\pi(x^3)^2)dx \\\pi\int\limits_{0}^{3}(81x^2-x^6)dx \\2\times \int\limits_{0}^{3}[\pi(9x)^2-\pi(x^3)^2]dx \\\pi\left[81\frac{x^3}{3}-\frac{x^7}{7}\right]_{0}^{3} \\ \\\frac{2916\pi}{7}$