Question

For each pair of definite integrals in Exercises 13–18, decide which, if either, is larger, without computing any integrals.

$\mathrm{\pi }{\int }_0^4\mathrm{ydy}\mathrm{and}\mathrm{\pi }{\int }_4^8(8-\mathrm{y})\mathrm{dy}$

Short answer

The value of both the integrals are equal.

Step-by-step solution

Step 1. Given Information

The given integrals are$\mathrm{\pi }{\int }_0^4\mathrm{ydy}\mathrm{and}\mathrm{\pi }{\int }_4^8(8-\mathrm{y})\mathrm{dy}$

Step 2. Explanation

The integrals are of the form $\mathrm{\pi }{\int }_a^b(f{\left(y\right))}^2\mathrm{dy}$which represents the volume of the solid formed by rotating the region bound by f(y)and y-axis in the interval$[a,b]$around x-axis.

On comparing the given integrals with the above given relation and observed that the function involved are $f\left(y\right)=\sqrt{y}\mathrm{in}\mathrm{the}\mathrm{interval}[0,4]\mathrm{and}f\left(y\right)=\sqrt{8-y}\mathrm{in}\mathrm{the}\mathrm{interval}[4,8]$

Now, we will plot the graph of both the functions and identify the regions between the interval.

From the graph, we can say that both the regions are identical and symmetrical. This means the volume of the solids formed by rotating these regions will be equal. Thus, the value of both the integrals is also equal.