Volumes of Revolution — Practice Problems and Worked Solutions

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16 worked problems

Every step is generated and checked by a computer algebra system.

1. 1.

   Find the volume when the region is revolved about the x-axis.

   $$R(x)=x,0\le x\le 1$$

   Answer: $\frac{\pi }{3}$

   Step-by-step solution

   1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
   2. $$V=\pi {\int }_0^1(x{)}^{2}dx=\frac{\pi }{3}$$
2. 2.

   Find the volume when the region is revolved about the x-axis.

   $$R(x)=x^2,0\le x\le 1$$

   Answer: $\frac{\pi }{5}$

   Step-by-step solution

   1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
   2. $$V=\pi {\int }_0^1(x^2{)}^{2}dx=\frac{\pi }{5}$$
3. 3.

   Find the volume when the region is revolved about the x-axis.

   $$R(x)=\sqrt{x},0\le x\le 4$$

   Answer: $8\pi$

   Step-by-step solution

   1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
   2. $$V=\pi {\int }_0^4(\sqrt{x}{)}^{2}dx=8\pi$$
4. 4.

   Find the volume when the region is revolved about the x-axis.

   $$R(x)=2x,0\le x\le 2$$

   Answer: $\frac{32\pi }{3}$

   Step-by-step solution

   1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
   2. $$V=\pi {\int }_0^2(2x{)}^{2}dx=\frac{32\pi }{3}$$
5. 5.

   Find the volume when the region is revolved about the x-axis.

   $$R(x)=x+1,0\le x\le 2$$

   Answer: $\frac{26\pi }{3}$

   Step-by-step solution

   1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
   2. $$V=\pi {\int }_0^2(x+1{)}^{2}dx=\frac{26\pi }{3}$$
6. 6.

   Find the volume when the region is revolved about the x-axis.

   $$R(x)=3-x,0\le x\le 3$$

   Answer: $9\pi$

   Step-by-step solution

   1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
   2. $$V=\pi {\int }_0^3(3-x{)}^{2}dx=9\pi$$
7. 7.

   Find the volume when the region is revolved about the x-axis.

   $$R(x)=x,0\le x\le 2$$

   Answer: $\frac{8\pi }{3}$

   Step-by-step solution

   1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
   2. $$V=\pi {\int }_0^2(x{)}^{2}dx=\frac{8\pi }{3}$$
8. 8.

   Find the volume when the region is revolved about the x-axis.

   $$R(x)=x^2,0\le x\le 2$$

   Answer: $\frac{32\pi }{5}$

   Step-by-step solution

   1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
   2. $$V=\pi {\int }_0^2(x^2{)}^{2}dx=\frac{32\pi }{5}$$
9. 9.

   Find the volume when the region is revolved about the x-axis.

   $$R(x)=3x,0\le x\le 1$$

   Answer: $3\pi$

   Step-by-step solution

   1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
   2. $$V=\pi {\int }_0^1(3x{)}^{2}dx=3\pi$$
10. 10.

    Find the volume when the region is revolved about the x-axis.

    $$R(x)=\sqrt{x},0\le x\le 9$$

    Answer: $\frac{81\pi }{2}$

    Step-by-step solution

    1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
    2. $$V=\pi {\int }_0^9(\sqrt{x}{)}^{2}dx=\frac{81\pi }{2}$$
11. 11.

    Find the volume when the region is revolved about the x-axis.

    $$R(x)=x+2,0\le x\le 1$$

    Answer: $\frac{19\pi }{3}$

    Step-by-step solution

    1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
    2. $$V=\pi {\int }_0^1(x+2{)}^{2}dx=\frac{19\pi }{3}$$
12. 12.

    Find the volume when the region is revolved about the x-axis.

    $$R(x)=2-x,0\le x\le 2$$

    Answer: $\frac{8\pi }{3}$

    Step-by-step solution

    1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
    2. $$V=\pi {\int }_0^2(2-x{)}^{2}dx=\frac{8\pi }{3}$$
13. 13.

    Find the volume when the region is revolved about the x-axis.

    $$R(x)=x^2,1\le x\le 2$$

    Answer: $\frac{31\pi }{5}$

    Step-by-step solution

    1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
    2. $$V=\pi {\int }_1^2(x^2{)}^{2}dx=\frac{31\pi }{5}$$
14. 14.

    Find the volume when the region is revolved about the x-axis.

    $$R(x)=4x,0\le x\le 1$$

    Answer: $\frac{16\pi }{3}$

    Step-by-step solution

    1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
    2. $$V=\pi {\int }_0^1(4x{)}^{2}dx=\frac{16\pi }{3}$$
15. 15.

    Find the volume when the region is revolved about the x-axis.

    $$R(x)=x+3,0\le x\le 2$$

    Answer: $\frac{98\pi }{3}$

    Step-by-step solution

    1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
    2. $$V=\pi {\int }_0^2(x+3{)}^{2}dx=\frac{98\pi }{3}$$
16. 16.

    Find the volume when the region is revolved about the x-axis.

    $$R(x)=\sqrt{x},1\le x\le 4$$

    Answer: $\frac{15\pi }{2}$

    Step-by-step solution

    1. $$V=\pi {\int }_a^b[R(x){]}^{2}dx$$
    2. $$V=\pi {\int }_1^4(\sqrt{x}{)}^{2}dx=\frac{15\pi }{2}$$

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