Petrus
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Calculate the volumes of the rotation bodies which arises when the area D in the xy-plane bounded by x-axis and curve $$7x-x^2$$may rotate around x- respective y-axes.
I will calculate $$V_x$$ and $$V_y$$ I start to get crit point $$x_1=0$$ and $$x_2=7$$
rotate on y-axe:
$$2\pi\int_a^bf(x)dx$$
so we get $$2\pi[\frac{7x^2}{2}-\frac{x^3}{3}]_0^7$$ $$V_y=\frac{2\pi*343}{6}$$
rotate on x axe:
$$\pi\int_a^bf(x)^2dx$$
so we start with:$$(7x-x^2)^2=49x^2-14x^3+x^4$$ so we get $$[\frac{49x^3}{3}-\frac{14x^4}{4}+\frac{x^5}{5}]_0^7$$ that means $$V_x=\frac{16087\pi}{30}$$ What I am doing wrong?
(Sorry for bad english.)
I will calculate $$V_x$$ and $$V_y$$ I start to get crit point $$x_1=0$$ and $$x_2=7$$
rotate on y-axe:
$$2\pi\int_a^bf(x)dx$$
so we get $$2\pi[\frac{7x^2}{2}-\frac{x^3}{3}]_0^7$$ $$V_y=\frac{2\pi*343}{6}$$
rotate on x axe:
$$\pi\int_a^bf(x)^2dx$$
so we start with:$$(7x-x^2)^2=49x^2-14x^3+x^4$$ so we get $$[\frac{49x^3}{3}-\frac{14x^4}{4}+\frac{x^5}{5}]_0^7$$ that means $$V_x=\frac{16087\pi}{30}$$ What I am doing wrong?
(Sorry for bad english.)