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The integral represents the volume of a solid. Describe the solid. $\pi\int\limits_0^1(y^4−y^8)\,dy$

  • The integral represents the volume of the solid obtained by rotating the region $R=\{\{x, y\}| 0\leq y\leq 1, y^4\leq x\leq y^2\}$of the $xy-$plane about the $x-$axis.
  • The integral represents the volume of the solid obtained by rotating the region $R=\{\{x, y\}| 0\leq y\leq 1, y^2\leq x\leq y^4\}$of the $xy-$plane about the $x-$axis.
  • The integral represents the volume of the solid obtained by rotating the region $R=\{\{x, y\}| 0\leq y\leq 1, y^4\leq x\leq y^2\}$ of the $xy-$plane about the $y-$axis.
  • The integral represents the volume of the solid obtained by rotating the region $R=\{\{x, y\}| 0\leq y\leq 1, y^2\leq x\leq y^4\}$ of the $xy-$plane about the $y-$axis.
  • The integral represents the volume of the solid obtained by rotating the region $R=\{\{x, y\}| 0\leq y\leq 1, y^4\leq x\leq y^8\}$ of the $xy-$plane about the $y-$axis.

This question aims to figure out the axis of rotation and the region within which the solid is bounded by using the given integral for the volume of the solid.

The volume of a solid is determined by rotating a region about a vertical or a horizontal line that does not pass through that plane. 

A washer is similar to a circular disk, but it has a hole in the center. This approach is used when indeed the axis of rotation is not the boundary of the region, and the cross-section is perpendicular to the axis of rotation. 

Expert Answer

Since the volume of a washer is calculated using both the inner radius $r_1 = \pi r^2$ and the outer radius $r_2=\pi R^2$ and is given by:

$V=\pi\int\limits_{a}^{b} (R^2 – r^2)\,dx$

The inner and outer radii of a washer will be written as functions of $x$ if it is perpendicular to the $x-$axis and the radii will be expressed as functions of $y$ if it is perpendicular to the $y-$axis.

Hence, the correct answer is (c)

Reason

Let  $V$ be the volume of the solid then

$V=\pi\int\limits_0^1(y^4−y^8)\,dy$

$V=\pi\int\limits_0^1[(y^2)^2−(y^4)^2]\,dy $

So, by washer method

Axis of rotation $=y-$axis

Upper boundary $x=y^2$

Lower boundary $x=y^4$

Therefore, the region is the $xy-$plane

$ y^4\leq x\leq y^2$

$0\leq y\leq 1$

Examples

Determine the volume $(V)$ of the solid generated by rotating the region bounded by the equations $y = x^2 +3$ and $y = x + 5$ about the $x-$axis.

Because $y = x^2 +3$ and $y = x +5$, we find that:

$x^2+3=x+5$

$x^2-x= -3+5$

$x^2-x-2=0$

$x^2-2x+x-2=0$

$(x-2)(x+1)=0$

$x=-1$ or $x=2$

So, the points of intersection of the graphs are $(-1,4)$ and  $(2,7)$

along with $x +5 \geq x^2 +3$ in the interval $[–1,2]$.

Geogebra export

And now using the washer method,

$V=\pi\int\limits_{-1}^{2}[(x+5)^2-(x^2+3)^2]\,dx$

$=\pi\int\limits_{-1}^{2}[(x^2+10x+25) -(x^4+6x^2+9)]\,dx$

$=\pi\int\limits_{-1}^{2}[-x^4-5x^2+10x+16]\,dx$

$=\pi\left[-\dfrac{x^5}{5}-\dfrac{5}{3}x^3+5x^2+16x\right]_{-1}^{2}\,dx$

$=\pi\left[-\dfrac{108}{5}+63\right]$

$V=\dfrac{207}{5}\,\pi$

 Images/mathematical drawings are created with GeoGebra.

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