Question

12. \(y = \sqrt x ,y = {x^2}\) \(\);about \(y = 2\)

Short answer

\(\int_0^1 \pi \left( {{{\left( {2 - {x^2}} \right)}^2} - {{(2 - \sqrt x )}^2}} \right)\)

Step-by-step solution

To Find the x-axis equation

1. Let's first notice that by rotating the graphed region around \(y = 2\)we will get a washer type solid. Therefore we use the follow it formulas:

\(\int_a^b A (x)dx\)

\(A(x) = \pi {( outer radius )^2} - \pi {( inner radius )^2}\)

2. We now have to find our inner and outer radius and our interval:

Inner radius: distance between\(y = \sqrt x \)and\(y = 2\). Thus we have:

\({r_1} = 2 - \sqrt x \)

Outer radius: distance between\(y = {x^2}\)and\(y = 2\). Thus we have:

\({r_2} = 2 - {x^2}\)

Interval: we can see from the picture(or easily calculate) that are curves intersect for\(x = 0\)and\(x = 1\).

3. Now we have everything needed so let us make the necessary substitutions:

\(\int_0^1 \pi {\left( {2 - {x^2}} \right)^2} - \pi {(2 - \sqrt x )^2}\;dx\)

\( = \)\(\int_0^1 \pi \left( {{{\left( {2 - {x^2}} \right)}^2} - {{(2 - \sqrt x )}^2}} \right)\)

Final Answer

The equation of the \(\int_0^1 \pi \left( {{{\left( {2 - {x^2}} \right)}^2} - {{(2 - \sqrt x )}^2}} \right)\)