Volume as double integrals

Question

Find the volume bounded by the cylinder $x^2$+$y^2$=4 and the planes y+z=4 and z=0.

My book tells: From the fig, it is self evident that z=4-y is to be integrated over the circle $x^2$+$y^2$=4 in the XY plane. To cover the shaded half of this circle, x varies from 0 to √(4-$y^2$) and y varies from -2 to 2.

And they set the integral : Volume=$$2\int_{-2}^2\int_0^\sqrt{4-y^2} (4-y)\;dx\;dy$$

Can someone tell me how did they set up this integration. Treat me as a beginner

Answer 1

The setp up is as follow

$$2\int_{-2}^2\int_0^\sqrt{4-y^2} (4-y)\;dx\;dy$$

indeed

• we are considering half of the volume starting from the base area of the cylinder
• $y$ varies from $-2$ to $2$
• $x$ varies form $0$ to $\sqrt{4-y^2}$
• the height above is $z=4-y$