Find the area of the indicated region.
Enclosed by $y = 4 x ^ { 4 } - 16 x ^ { 2 }$ and $y = 16 x ^ { 2 } - 4 x ^ { 4 }$ .

A) $\frac { 256 } { 15 }$
B) 16
C) $\frac { 15 } { 1,024 }$
D) $\frac { 1,024 } { 15 }$
E) 4

Question

Find the area of the indicated region.  
Enclosed by $y = 4 x ^ { 4 } - 16 x ^ { 2 }$ and $y = 16 x ^ { 2 } - 4 x ^ { 4 }$ .
 
A) $\frac { 256 } { 15 }$
B) 16
C)   $\frac { 15 } { 1,024 }$
D) $\frac { 1,024 } { 15 }$
E) 4

Quizplus · Accepted Answer

To find the area between the curves $y = 4x^4 - 16x^2$ and $y = 16x^2 - 4x^4$, we set them equal to each other to find the intersection points. However, since they are symmetric with respect to the y-axis and are opposites of each other, we can integrate the difference between the two functions from 0 to the first positive root to find the area of one section and then double it to get the total area. The difference between the two functions is $|16x^2 - 4x^4 - (4x^4 - 16x^2)| = 32x^2 - 8x^4$. Integrating this from 0 to 1 (since the roots are at x=0 and x=1) gives us the area of one lobe. The integral of $32x^2 - 8x^4$ from 0 to 1 is $\frac{32}{3} - \frac{8}{5}$, which simplifies to $\frac{128}{15}$. Doubling this result for both sides of the y-axis gives us $\frac{256}{15}$, which corresponds to choice D.

Find the Area of the Indicated Region y=4x4−16x2y = 4 x ^ { 4 } - 16 x ^ { 2 }y=4x4−16x2

Find the Area of the Indicated Region $y = 4 x ^ { 4 } - 16 x ^ { 2 }$