How do you find the volume bounded by #x^2y^2+16y^2=6# and the x & y axes, the line x=4 revolved about the x-axis?

Question

Charles Anctil · Accepted Answer

#3/8pi^2#

= 3.7011 cubic units, nearly.

The graph of #x^2y^2+16y^2=6# is inserted. In #Q_1#, The curve meets x = 0 ( y-axis ) at #(0, 1/2sqrt(3/2))# and the parallel x = 4 at #( 4, sqrt3/4 ).# The volume is #V = pi int y^2 dx#, from x = 0 to x = 4 #=pi int 6/(x^2+4^2) dx#, between the limits #=6pi(1/4)[tan^(-1)(x/4)]#, between the limits #=3/2pi[tan^(-1)1-tan^(-1)0]# #3/2pi(pi/4)# #=3/8pi^2# = 3.7011 cubic units, nearly.- graph{x^2y^2+16y^2-6=0 [-10, 10, -5, 5]}

Paisley Johnson · Answer

undefined To find the volume bounded by $ x^2y^2 + 16y^2 = 6 $ and the x and y axes, with the line $ x = 4 $ revolved about the x-axis, you can use the method of cylindrical shells.

The formula for the volume of a solid generated by rotating a curve $ y = f(x) $ from $ x = a $ to $ x = b $ about the x-axis using cylindrical shells is:

$$ V = \int_{a}^{b} 2\pi x \cdot f(x) \, dx $$

In this case, we need to express $ y $ in terms of $ x $ from the given equation $ x^2y^2 + 16y^2 = 6 $, which simplifies to $ y = \frac{\sqrt{6}}{\sqrt{x^2 + 16}} $.

The bounds of integration $ a $ and $ b $ can be determined by finding the x-intercepts of the curve, which are the points where it intersects the x-axis. Solving $ x^2y^2 + 16y^2 = 6 $ when $ y = 0 $ gives us $ x^2 = 6 $, so $ x = \pm \sqrt{6} $. We'll integrate from $ x = -\sqrt{6} $ to $ x = 4 $.

Now, we'll plug $ f(x) = \frac{\sqrt{6}}{\sqrt{x^2 + 16}} $ into the formula and solve the integral:

$$ V = \int_{-\sqrt{6}}^{4} 2\pi x \cdot \frac{\sqrt{6}}{\sqrt{x^2 + 16}} \, dx $$

This integral might be challenging to solve directly. However, recognizing that the integrand is even, we can integrate from $ x = 0 $ to $ x = 4 $ and double the result:

$$ V = 2 \int_{0}^{4} 2\pi x \cdot \frac{\sqrt{6}}{\sqrt{x^2 + 16}} \, dx $$

$$ = 4\pi \int_{0}^{4} \frac{x\sqrt{6}}{\sqrt{x^2 + 16}} \, dx $$

This integral might require a trigonometric substitution. Once solved, it will give you the volume of the solid of revolution.

Therefore, the volume bounded by $ x^2y^2 + 16y^2 = 6 $ and the x and y axes, with the line $ x = 4 $ revolved about the x-axis, can be found using the cylindrical shells method and the definite integral as described above.

How do you find the volume bounded by #x^2y^2+16y^2=6# and the x &amp; y axes, the line x=4 revolved about the x-axis?

How do you find the volume bounded by #x^2y^2+16y^2=6# and the x & y axes, the line x=4 revolved about the x-axis?