How do you find the volume of the solid obtained by rotating the region bounded by the curves #x=0# and #Y=4-x^2# and #y=3x# rotated around the y-axis?

Question

Ezra Adams · Accepted Answer

#(158pi)/15#

The cross-sections (head on views of vertical slices) for this solid of revolution will be "washers" (the more official name is annuli). The areas of the cross-sections will be given by #pi r_{o}^{2}-pi r_{i}^{2}#, where #r_{i}# is the inner radius and #r_{o}# is the outer radius.

Based on a picture of the region you are revolving around the #x#-axis, the inner and outer radii are the following functions of #x#: #r_{i}=3x# and #r_{o}=4-x^2#.

Therefore, the cross-sectional area, as a function of #x#, is

#A(x)=pi((4-x^2)^2-(3x)^2)=pi(16-17x^2+x^4)#

The volume of the resulting solid is given as a definite integral of #A(x)# over an appropriate interval. Looking at your picture, the lower limit of the integral should be #x=0#. To find the upper limit, solve #4-x^2=3x# to get #x^2+3x-4=0# so #(x+4)(x-1)=0#, which has #x=1# as its only positive root, making that the upper limit of the integral.

Hence

#V=int_{0}^{1}A(x)\ dx=pi int_{0}^{1}(16-17x^2+x^4)\ dx#

#=pi (16x- 17/3 x^3+ 1/5 x^5)|_{0}^{1}=pi(16-17/3+1/5)#

#=pi((240-85+3)/15)=(158pi)/15#

Here's a picture of the 2-dimensional region that gets revolved around the #x#-axis:

And here's a picture of the 3-dimensional solid (with a different scale on the #x#-axis so it looks more stretched out). The solid itself is the part "between" the red and the blue (BTW, I made the sides of the solid out of circles with space between them just so you could see inside the solid...in reality, the space between the circles is part of the sides of the solid as well).

Christopher Allers · Answer

undefined The volume of the solid obtained by rotating the region bounded by the curves $x = 0$, $y = 4 - x^2$, and $y = 3x$ around the y-axis can be found using the method of cylindrical shells. The formula for the volume of the solid obtained by rotating a region around the y-axis using cylindrical shells is:

$$ V = 2\pi \int_a^b x \cdot h(x) \, dx $$

where $a$ and $b$ are the x-values of the intersection points of the curves, and $h(x)$ is the height of the shell at the x-value $x$.

In this case, the intersection points of the curves are $x = 0$ and $y = 4 - x^2$, so we need to find the x-values where $y = 4 - x^2$ intersects with $y = 3x$.

Setting $4 - x^2 = 3x$ and solving for $x$ gives $x^2 + 3x - 4 = 0$, which factors to $(x + 4)(x - 1) = 0$. So, the intersection points are $x = -4$ and $x = 1$.

The height of the shell $h(x)$ is the difference between the y-values of the curves at the x-value $x$, which is $h(x) = 4 - x^2 - 3x$.

Now, we can calculate the volume using the formula:

$$ V = 2\pi \int_{-4}^1 x \cdot (4 - x^2 - 3x) \, dx $$

$$ V = 2\pi \int_{-4}^1 (4x - x^3 - 3x^2) \, dx $$

$$ V = 2\pi \left[ 2x^2 - \frac{x^4}{4} - x^3 \right]_{-4}^1 $$

$$ V = 2\pi \left[ (2(1)^2 - \frac{(1)^4}{4} - (1)^3) - (2(-4)^2 - \frac{(-4)^4}{4} - (-4)^3) \right] $$

$$ V = 2\pi \left[ (2 - \frac{1}{4} - 1) - (32 - 64 - (-64)) \right] $$

$$ V = 2\pi \left[ \frac{7}{4} - (-32) \right] $$

$$ V = 2\pi \left[ \frac{7}{4} + 32 \right] $$

$$ V = 2\pi \cdot \frac{135}{4} $$

$$ V = \frac{270\pi}{4} $$

$$ V = \frac{135\pi}{2} $$

So, the volume of the solid obtained by rotating the region around the y-axis is $ \frac{135\pi}{2} $.