# Integration help

I tried to find the volume of a torus revolved around the y-axis using the washer method. The equation for the circle is (x-2)^2 +y^2=1.
I need to integrate (1-y^2)^1/2 dy from 1 to 0 multiplied by a constant 16pi. The constant is not really important but I can't seem to integrate the integrand. I am aware that there's a formula for finding a vol. of a torus, as well.

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## Answers and Replies

mathman
Science Advisor
Unless you misstated the problem, it is trivial. The integral of 1 is y and the integral of y2 is y3/3. Just put them together.

Yes, but in this case, the (1-y^2) are under the square root, you can't separate the two terms.

$$\int_0^1 \sqrt (1-y^{2}) dy$$
a trig substitution would work here, like y=sinx, so dy=cos(x)dx, when y=0, x=0, when y=1,$$x=\frac{\pi}{2}$$

$$\int_0^\frac{\pi}{2} \sqrt(1-sin^{2}x )cosxdx$$ now remember

$$1-sin^{2}(x)=cos^{2}(x)$$ so

$$\int_0^\frac{\pi}{2} cos^{2}(x) dx=\int_0^\frac{\pi}{2} \frac{1+cos(2x)}{2} dx$$ and you are almost done!!

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