0
$\begingroup$

Find the volume of the solid in the first quadrant bounded by the coordinate planes, the cylinder $x^{2} + y^{2}=4$, and the plane $z+y=3$.


If we draw the graph, then the integral will be calculated should be

$$ \int_{0}^{2} \int_{0}^{\sqrt{4-x^{2}}} (3-y) \: dy dx $$

with $3-y = z = f(x,y)$.

The boundary $ \sqrt{4-x^{2}} $ is from the cylinder. Is this correct? Thanks.

$\endgroup$
1
$\begingroup$

Yes, that is correct. And if you are going to evaluate the integral, I suggest polar coordinates.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.