Question

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. Y = (4/9) x^2, y = (13/9) − x2; about the x-axis

Answer 1

Answer:

V = 8.06 cubed units

Step-by-step explanation:

You have the following curves:

$y_1=\frac{4}{9}x^2=f(x)\\\\y_2=\frac{13}{9}-x^2=g(x)$

In order to calculate the solid of revolution bounded by the previous curves and the x axis, you use the following formula:

$V=\pi \int_a^b [(g(x))^2-(f(x))^2]dx$       (1)

To determine the limits of the integral you equal both curves f=g and solve for x:

$f(x)=g(x)\\\\\frac{4}{9}x^2=\frac{13}{9}-x^2\\\\\frac{4}{9}x^2+x^2=\frac{13}{9}\\\\\frac{13}{9}x^2=\frac{13}{9}\\\\x=\pm 1$

Then, the limits are a = -1 and b = 1

You replace f(x), g(x), a and b in the equation (1):

$V=\pi \int_{-1}^{1}[(\frac{13}{9}-x^2)^2-(\frac{4}{9}x^2)^2]dx\\\\V=\pi \int_{-1}^1[\frac{169}{81}-\frac{26}{9}x^2+x^4-\frac{16}{81}x^4]dx\\\\V=\pi \int_{-1}^1 [\frac{169}{81}-\frac{26}{9}x^2+\frac{65}{81}x^4]dx\\\\V=\pi [\frac{169}{81}x-\frac{26}{27}x^3+\frac{65}{405}x^5]_{-1}^1\\\\V\approx8.06\ cubed\ units$

The volume of the solid of revolution is approximately 8.06 cubed units