Question

Consider the region bounded by y = ex, y = 4, and x = 0 . A solid is created so that the given region is its base and cross-sections perpendicular to the y-axis are squares. Set up a Riemann sum and then a definite integral needed to find the volume of the solid. What is the approximate volume of a slice perpendicular to the y-axis? You can type the word Delta or use the CalcPad when you need Δ. (ln(y)2)Δy In your definite integral what is the lower endpoint? 0 In your definite integral what is the upper endpoint? 1 Evaluate the integral. Give an exact answer.

Answer 1

Each cross section has side length equal to $x$ satisfying $y=e^x\implies x=\ln y$, where $0\le x\le\ln4$ so that $1\le y\le4$.

The exact volume is given by the definite integral,

$\displaystyle\int_1^4(\ln y)^2\,\mathrm dy$

Take a slice at any value of $y$ with thickness $\Delta y$. Then the slice has volume $(\ln y)^2\Delta y$.

The approximate total volume of these slices is then given by the Riemann sum,

$\displaystyle\sum_{i=1}^n(\ln y_i)^2\Delta y_i$

where $y_i$ are chosen however you like from the range above.

Compute the definite integral above for the exact volume: you can do this by parts, taking

$u=(\ln y)^2\implies\mathrm du=\dfrac{2\ln y}y\,\mathrm dy$

$\mathrm dv=\mathrm dy\implies v=y$

$\implies\displaystyle\int_1^4(\ln y)^2\,\mathrm dy=y(\ln y)^2\bigg|_1^4-2\int_1^4\ln y\,\mathrm dy$

The remaining integral can be done by parts again, this time with

$u=\ln y\implies\mathrm du=\dfrac{\mathrm dy}y$

$\mathrm dv=\mathrm dy\implies v=y$

$\implies\displaystyle\int_1^4\ln y\,\mathrm dy=y\ln y\bigg|_1^4-\int_1^4\mathrm dy$

and of course

$\displaystyle\int_1^4\mathrm dy=y\bigg|_1^4$

So we have

$\displaystyle\int_1^4(\ln y)^2\,\mathrm dy=4(\ln 4)^2-2(4\ln 4-(4-1))$

$\displaystyle\int_1^4(\ln y)^2\,\mathrm dy=4(\ln 4)^2-8\ln 4+6$