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Question Number 197027 by sonukgindia last updated on 06/Sep/23

Answered by MM42 last updated on 06/Sep/23

$$c_1:x^2+y^2=36\mathrm{\&}{c}_2:{(x-10)}^2+y^2=36$$$$c_1,c_2\Rightarrow A=5,B=6$$$$v_1=\frac{1}{2}\times \frac{4}{3}\pi \times 6^3=144\pi$$$$v_2=\pi {\int }_0^5(36-x^2)dx=\pi (36x-\frac{x^3}{3}){\mid }_0^5=\frac{415\pi }{3}$$$$v_3=\pi {\int }_5^6(36-{(x^2-10)}^2)dx=\pi (36x-\frac{{(x-10)}^3}{3}){\mid }_5^6=\frac{47\pi }{3}$$$$v=307\pi ✓$$$$$$

Answered by MM42 last updated on 06/Sep/23

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