8.2.2 Int area of a region and graph

[karush]

a. Sketch the region of integration and evaluate the Integral
b. Evaluate
$V=5\displaystyle\int_0^8 \biggr[ x^3\biggr]_{(y-4)/2}^{y^{1/3}}\ dy \
=5\displaystyle\int_0^8 [(y^{1/3})^3-((y-4)/2))^3] \quad \ dy \
=5\displaystyle\int_0^8 \biggr [y-\dfrac{(y-4)^3}{8}\biggr] \ dy$
Expand $5\displaystyle\int_0^8 -\dfrac{y^3}{8}+\dfrac{3y^3}{2}-5y +8 \ dy
=5\biggr[-\dfrac{y^4}{32}+\dfrac{y^3}{2}-\dfrac{5y^2}{2}+8y \biggr]_0^8
=\biggr[-\dfrac{8^4}{32}+\dfrac{8^3}{2}-\dfrac{5(8)^2}{2}+8(8) \biggr]=160$

ok there is bound to be some typos in this or too much content
want to make sure this is correct before i attempt a tikz of the region
mahalo for suggestions and such

 

[jvicens]

a. The region of integration can be sketched as follows:

The integral is being taken over the region between the curves $x=\left(\dfrac{y-4}{2}\right)^3$ and $x=y^{1/3}$, bounded by the lines $y=0$ and $y=8$. This region can be represented as the shaded area in the graph below:

b. To evaluate the integral, we can first rewrite it as:

$V=5\displaystyle\int_0^8 [(y^{1/3})^3-((y-4)/2))^3] \ dy \
=5\displaystyle\int_0^8 \biggr [y-\dfrac{(y-4)^3}{8}\biggr] \ dy$

Expanding this, we get:

$V=5\displaystyle\int_0^8 -\dfrac{y^3}{8}+\dfrac{3y^3}{2}-5y +8 \ dy$

Using the power rule for integration, we can evaluate this integral to get:

$V=5\biggr[-\dfrac{y^4}{32}+\dfrac{y^3}{2}-\dfrac{5y^2}{2}+8y \biggr]_0^8$

Substituting the upper and lower limits of integration, we get:

$V=\biggr[-\dfrac{8^4}{32}+\dfrac{8^3}{2}-\dfrac{5(8)^2}{2}+8(8) \biggr]$

Simplifying, we get the final answer:

$V=160$ units$^3$

Overall, the integral evaluates to the volume of the region bounded by the given curves and lines. This can also be verified by using a triple integral in cylindrical coordinates, but the method used above is simpler and more straightforward.