Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

8. \(y = {x^2} + 1,y = 9 - {x^2},\;\)about \(y = - 1\)

Short answer

The volume of the solid obtained is \(256\pi \)

Step-by-step solution

Find the Intersection

\(f(x) = {x^2} + 1,g(x) = 9 - {x^2},\;\;\;\)about \(y = - 1\)

Find the intersections

\({x^2} + 1{\rm{ }} = 9 - {x^2}\)

\(2{x^2}{\rm{ }} = 8\)

\({x^2}{\rm{ }} = 4\)

\(x = \pm 2\)

We will use symmetry to integrate from \(0\) to \(2\) and multiply the integral by \(2\).

\(y = - 1\)is below both curves, so subtract from each curve to get the vertical distances which will be the radii of the washers. (subtract a lesser value from a greater value: \({r_1} \ge {r_2})\)

\({r_1}{\rm{ }} = 9 - {x^2} - ( - 1)\)

\( = 10 - {x^2}\)

\({r_2}{\rm{ }} = {x^2} + 1 - ( - 1)\)

\( = {x^2} + 2\)

To find the graph curves

Integrating with respect to using the\(x{\rm{ }}\) washer method will be the simplest way to do this problem.

An example washer at \(x = 1{\rm{ }}\)is shown.

Final proof

Integrate subtracting the lower from the upper curve, since we want a positive answer. (Subtract a lesser value from a greater value: \({r_1} \ge {r_2}\))

\(V = \int_a^b \pi \left( {r_1^2 - r_2^2} \right)dx{\rm{ }} = 2\pi \int_0^2 {\left( {{{\left( {10 - {x^2}} \right)}^2} - {{\left( {{x^2} + 2} \right)}^2}} \right)} dx\)

\( = 2\pi \int_0^2 {\left( {100 - 20{x^2} + {x^4} - {x^4} - 4{x^2} - 4} \right)} dx\)

\( = 2\pi \int_0^2 {\left( {96 - 24{x^2}} \right)} dx\)

\( = 2 \cdot 24\pi \int_0^2 {\left( {4 - {x^2}} \right)} dx\)

\( = 48\pi \left( {4x - \frac{1}{3}{x^3}} \right)_0^2\)

\( = 48\pi \left( {4(2) - \frac{1}{3}{{(2)}^3} - (0 - 0)} \right)\)

\( = 48\pi \left( {8 - \frac{8}{3}} \right)\)

\( = 48\pi \left( {\frac{{24 - 8}}{3}} \right)\)

\( = 48\pi \left( {\frac{{16}}{3}} \right)\)

\( = 16\pi (16)\)

\( = 256\pi \)