Find the volume of the solid whose base is bounded by $y=x$ and $y=x^3-3 x$, having cross sections that are right triangles with a height of 4, taken perpendicular to the $x$-axis.
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Answer (2)
Find the intersection points of y = x and y = x 3 − 3 x , which are x = − 2 , 0 , 2 .
Determine the base of the right triangle as ∣4 x − x 3 ∣ .
Calculate the area of the right triangle cross section: A ( x ) = 2∣4 x − x 3 ∣ .
Integrate the area over the interval [ − 2 , 2 ] : V = ∫ − 2 2 2∣4 x − x 3 ∣ d x = 16 .
The volume of the solid is 16 .
Explanation
Problem Setup We are asked to find the volume of the solid whose base is bounded by y = x and y = x 3 − 3 x , and the cross sections perpendicular to the x -axis are right triangles with height 4. The volume can be found by integrating the area of the cross sections along the x -axis.
Finding Intersection Points First, we need to find the intersection points of the curves y = x and y = x 3 − 3 x . We set x = x 3 − 3 x , which gives x 3 − 4 x = 0 . Factoring out x , we have x ( x 2 − 4 ) = 0 , so x ( x − 2 ) ( x + 2 ) = 0 . The intersection points are x = − 2 , 0 , 2 . These will be our limits of integration.
Determining the Base of the Triangle Next, we need to determine which curve is above the other in each interval.
For − 2 < x < 0 , let's test x = − 1 . We have y = x = − 1 and y = x 3 − 3 x = ( − 1 ) 3 − 3 ( − 1 ) = − 1 + 3 = 2 . So x"> x 3 − 3 x > x in this interval.
For 0 < x < 2 , let's test x = 1 . We have y = x = 1 and y = x 3 − 3 x = ( 1 ) 3 − 3 ( 1 ) = 1 − 3 = − 2 . So x^3 - 3x"> x > x 3 − 3 x in this interval.
Thus, the base of the right triangle is given by ∣ x − ( x 3 − 3 x ) ∣ = ∣4 x − x 3 ∣ .
Area of the Triangle The area of the right triangle is A ( x ) = 2 1 ⋅ b ( x ) ⋅ h = 2 1 ∣4 x − x 3 ∣ ⋅ 4 = 2∣4 x − x 3 ∣ .
Calculating the Volume Now we can set up the integral for the volume. Since the expression inside the absolute value changes sign, we need to split the integral into intervals where 4 x − x 3 is positive or negative.
V = ∫ − 2 2 2∣4 x − x 3 ∣ d x = ∫ − 2 0 2 ( x 3 − 4 x ) d x + ∫ 0 2 2 ( 4 x − x 3 ) d x
Let's evaluate the first integral: ∫ − 2 0 2 ( x 3 − 4 x ) d x = 2 ∫ − 2 0 ( x 3 − 4 x ) d x = 2 [ 4 x 4 − 2 x 2 ] − 2 0 = 2 [( 0 ) − ( 4 ( − 2 ) 4 − 2 ( − 2 ) 2 )] = 2 [ 0 − ( 4 16 − 8 )] = 2 [ 0 − ( 4 − 8 )] = 2 [ 0 − ( − 4 )] = 2 ( 4 ) = 8
Now let's evaluate the second integral: ∫ 0 2 2 ( 4 x − x 3 ) d x = 2 ∫ 0 2 ( 4 x − x 3 ) d x = 2 [ 2 x 2 − 4 x 4 ] 0 2 = 2 [( 2 ( 2 ) 2 − 4 ( 2 ) 4 ) − ( 0 )] = 2 [( 8 − 4 16 )] = 2 [( 8 − 4 )] = 2 ( 4 ) = 8
So, the total volume is V = 8 + 8 = 16 .
Final Answer The volume of the solid is 16.
Examples
Imagine you are designing a uniquely shaped cake pan. The base of the pan is defined by the curves y = x and y = x 3 − 3 x . To determine how much batter you need, you calculate the volume of the pan by considering the cross-sections perpendicular to the x-axis as right triangles. This calculation ensures you use the right amount of batter, preventing overflow or underfilling, and resulting in a perfectly baked cake with a unique shape.
The volume of the solid with a base bounded by the curves y = x and y = x 3 − 3 x , having right triangular cross sections of height 4, is calculated as 16 . This is determined by integrating the area of the cross sections along the defined limits of intersection. The rigorous evaluation shows that the total volume amounts to 16 .
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